How to Integrate Using the Beta Function

Deriving the Beta Function

Begin with the product of two Gamma functions. This product is the first step into deriving the standard integral representation of the Beta function. Γ ( α ) Γ ( β ) = ∫ 0 ∞ x α − 1 e − x d x ∫ 0 ∞ y β − 1 e − y d y = ∫ 0 ∞ d x ∫ 0 ∞ d y x α − 1 y β − 1 e − x − y {\displaystyle {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{0}^{\infty }x^{\alpha -1}e^{-x}\mathrm {d} x\int _{0}^{\infty }y^{\beta -1}e^{-y}\mathrm {d} y\\&=\int _{0}^{\infty }\mathrm {d} x\int _{0}^{\infty }\mathrm {d} y\,x^{\alpha -1}y^{\beta -1}e^{-x-y}\end{aligned}}} {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{{0}}^{{\infty }}x^{{\alpha -1}}e^{{-x}}{\mathrm {d}}x\int _{{0}}^{{\infty }}y^{{\beta -1}}e^{{-y}}{\mathrm {d}}y\\&=\int _{{0}}^{{\infty }}{\mathrm {d}}x\int _{{0}}^{{\infty }}{\mathrm {d}}y\,x^{{\alpha -1}}y^{{\beta -1}}e^{{-x-y}}\end{aligned}}

Make the u-substitution u = x + y {\displaystyle u=x+y} u=x+y. We rewrite the double integral in terms of u {\displaystyle u} u and x . {\displaystyle x.} x. Γ ( α ) Γ ( β ) = ∫ 0 ∞ d x ∫ 0 ∞ d y x α − 1 y β − 1 e − x − y = ∫ 0 ∞ d u ∫ 0 u d x x α − 1 ( u − x ) β − 1 e − u = ∫ 0 ∞ d u ∫ 0 u d x x α − 1 u β − 1 ( 1 − x u ) β − 1 e − u {\displaystyle {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{0}^{\infty }\mathrm {d} x\int _{0}^{\infty }\mathrm {d} y\,x^{\alpha -1}y^{\beta -1}e^{-x-y}\\&=\int _{0}^{\infty }\mathrm {d} u\int _{0}^{u}\mathrm {d} x\,x^{\alpha -1}(u-x)^{\beta -1}e^{-u}\\&=\int _{0}^{\infty }\mathrm {d} u\int _{0}^{u}\mathrm {d} x\,x^{\alpha -1}u^{\beta -1}\left(1-{\frac {x}{u}}\right)^{\beta -1}e^{-u}\end{aligned}}} {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{{0}}^{{\infty }}{\mathrm {d}}x\int _{{0}}^{{\infty }}{\mathrm {d}}y\,x^{{\alpha -1}}y^{{\beta -1}}e^{{-x-y}}\\&=\int _{{0}}^{{\infty }}{\mathrm {d}}u\int _{{0}}^{{u}}{\mathrm {d}}x\,x^{{\alpha -1}}(u-x)^{{\beta -1}}e^{{-u}}\\&=\int _{{0}}^{{\infty }}{\mathrm {d}}u\int _{{0}}^{{u}}{\mathrm {d}}x\,x^{{\alpha -1}}u^{{\beta -1}}\left(1-{\frac {x}{u}}\right)^{{\beta -1}}e^{{-u}}\end{aligned}}

Make the u-sub t = x / u {\displaystyle t=x/u} t=x/u. Rewrite the double integral in terms of u {\displaystyle u} u and t . {\displaystyle t.} t. Now we see that the first integral is simply Γ ( α + β ) . {\displaystyle \Gamma (\alpha +\beta ).} \Gamma (\alpha +\beta ). Γ ( α ) Γ ( β ) = ∫ 0 ∞ d u ∫ 0 u d x x α − 1 u β − 1 ( 1 − x u ) β − 1 e − u = ∫ 0 ∞ d u u α + β − 1 e − u ∫ 0 1 d t t α − 1 ( 1 − t ) β − 1 = Γ ( α + β ) ∫ 0 1 t α − 1 ( 1 − t ) β − 1 d t {\displaystyle {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{0}^{\infty }\mathrm {d} u\int _{0}^{u}\mathrm {d} x\,x^{\alpha -1}u^{\beta -1}\left(1-{\frac {x}{u}}\right)^{\beta -1}e^{-u}\\&=\int _{0}^{\infty }\mathrm {d} u\,u^{\alpha +\beta -1}e^{-u}\int _{0}^{1}\mathrm {d} t\,t^{\alpha -1}(1-t)^{\beta -1}\\&=\Gamma (\alpha +\beta )\int _{0}^{1}t^{\alpha -1}(1-t)^{\beta -1}\mathrm {d} t\end{aligned}}} {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=\int _{{0}}^{{\infty }}{\mathrm {d}}u\int _{{0}}^{{u}}{\mathrm {d}}x\,x^{{\alpha -1}}u^{{\beta -1}}\left(1-{\frac {x}{u}}\right)^{{\beta -1}}e^{{-u}}\\&=\int _{{0}}^{{\infty }}{\mathrm {d}}u\,u^{{\alpha +\beta -1}}e^{{-u}}\int _{{0}}^{{1}}{\mathrm {d}}t\,t^{{\alpha -1}}(1-t)^{{\beta -1}}\\&=\Gamma (\alpha +\beta )\int _{{0}}^{{1}}t^{{\alpha -1}}(1-t)^{{\beta -1}}{\mathrm {d}}t\end{aligned}} Γ ( α ) Γ ( β ) Γ ( α + β ) = ∫ 0 1 t α − 1 ( 1 − t ) β − 1 d t {\displaystyle {\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}=\int _{0}^{1}t^{\alpha -1}(1-t)^{\beta -1}\mathrm {d} t} {\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}=\int _{{0}}^{{1}}t^{{\alpha -1}}(1-t)^{{\beta -1}}{\mathrm {d}}t Below, we go through three examples that make direct use of the Beta function.

Evaluate the integral below. ∫ 0 1 x 3 x ( 1 − x ) d x {\displaystyle \int _{0}^{1}x^{3}{\sqrt {x(1-x)}}\mathrm {d} x} \int _{{0}}^{{1}}x^{{3}}{\sqrt {x(1-x)}}{\mathrm {d}}x

Find α {\displaystyle \alpha } \alpha and β {\displaystyle \beta } \beta and substitute those values into the definition. We see that α = 9 / 2 {\displaystyle \alpha =9/2} \alpha =9/2 and β = 3 / 2 {\displaystyle \beta =3/2} \beta =3/2 just from inspection. ∫ 0 1 x 3 x ( 1 − x ) d x = Γ ( 9 / 2 ) Γ ( 3 / 2 ) Γ ( 6 ) {\displaystyle \int _{0}^{1}x^{3}{\sqrt {x(1-x)}}\mathrm {d} x={\frac {\Gamma (9/2)\Gamma (3/2)}{\Gamma (6)}}} \int _{{0}}^{{1}}x^{{3}}{\sqrt {x(1-x)}}{\mathrm {d}}x={\frac {\Gamma (9/2)\Gamma (3/2)}{\Gamma (6)}}

Simplify. Use the recursion relation to write the numerator in terms of π . {\displaystyle \pi .} \pi . Γ ( 9 / 2 ) Γ ( 3 / 2 ) Γ ( 6 ) = 1 120 7 2 5 2 3 2 π 4 = 7 π 256 {\displaystyle {\frac {\Gamma (9/2)\Gamma (3/2)}{\Gamma (6)}}={\frac {1}{120}}{\frac {7}{2}}{\frac {5}{2}}{\frac {3}{2}}{\frac {\pi }{4}}={\frac {7\pi }{256}}} {\frac {\Gamma (9/2)\Gamma (3/2)}{\Gamma (6)}}={\frac {1}{120}}{\frac {7}{2}}{\frac {5}{2}}{\frac {3}{2}}{\frac {\pi }{4}}={\frac {7\pi }{256}}

Evaluate the integral below. We see that our integrand is not quite in the form that we want it to, but we can take advantage of the fact that α {\displaystyle \alpha } \alpha and β {\displaystyle \beta } \beta are arbitrary parameters. ∫ 0 1 x 2 ( 1 − x 4 ) 3 d x {\displaystyle \int _{0}^{1}{\sqrt[{3}]{x^{2}(1-x^{4})}}\mathrm {d} x} \int _{{0}}^{{1}}{\sqrt[ {3}]{x^{{2}}(1-x^{{4}})}}{\mathrm {d}}x

Make the u-sub u = x 4 {\displaystyle u=x^{4}} u=x^{{4}}. This gets the quantity inside the parentheses into the form that we want. We changed the exponent on the power term, but since α {\displaystyle \alpha } \alpha is arbitrary, we don't have to worry. ∫ 0 1 x 2 ( 1 − x 4 ) 3 d x = 1 4 ∫ 0 1 u − 7 / 12 ( 1 − u ) 1 / 3 d u {\displaystyle \int _{0}^{1}{\sqrt[{3}]{x^{2}(1-x^{4})}}\mathrm {d} x={\frac {1}{4}}\int _{0}^{1}u^{-7/12}(1-u)^{1/3}\mathrm {d} u} \int _{{0}}^{{1}}{\sqrt[ {3}]{x^{{2}}(1-x^{{4}})}}{\mathrm {d}}x={\frac {1}{4}}\int _{{0}}^{{1}}u^{{-7/12}}(1-u)^{{1/3}}{\mathrm {d}}u

Evaluate using the Beta function. Simplify using the recursion relation to get the arguments of the Gamma functions between 0 and 1. Make sure your arithmetic skills are up to par. 1 4 ∫ 0 1 u − 7 / 12 ( 1 − u ) 1 / 3 d u = Γ ( 5 / 12 ) Γ ( 4 / 3 ) 4 Γ ( 7 / 4 ) = Γ ( 5 / 12 ) Γ ( 1 / 3 ) 9 Γ ( 3 / 4 ) {\displaystyle {\frac {1}{4}}\int _{0}^{1}u^{-7/12}(1-u)^{1/3}\mathrm {d} u={\frac {\Gamma (5/12)\Gamma (4/3)}{4\,\Gamma (7/4)}}={\frac {\Gamma (5/12)\Gamma (1/3)}{9\,\Gamma (3/4)}}} {\frac {1}{4}}\int _{{0}}^{{1}}u^{{-7/12}}(1-u)^{{1/3}}{\mathrm {d}}u={\frac {\Gamma (5/12)\Gamma (4/3)}{4\,\Gamma (7/4)}}={\frac {\Gamma (5/12)\Gamma (1/3)}{9\,\Gamma (3/4)}}

Evaluate the integral below. Of course, the Beta function can also directly be used to evaluate these types of integrals with logs attached to them. ∫ 0 1 x ( 1 − x ) 2 ln ⁡ x d x {\displaystyle \int _{0}^{1}x(1-x)^{2}\ln x\mathrm {d} x} \int _{{0}}^{{1}}x(1-x)^{{2}}\ln x{\mathrm {d}}x

Consider the integral below instead. This is standard procedure for an integral like this. We rewrite the power term so that e {\displaystyle e} e is in the base and expand that into its Taylor series. Then we find the appropriate coefficient, neglecting higher-order terms because ϵ {\displaystyle \epsilon } \epsilon is small (and therefore they go to 0 faster). ∫ 0 1 x 1 + ϵ ( 1 − x ) 2 d x = ∑ n = 0 ∞ ϵ n n ! ∫ 0 1 x ( 1 − x ) 2 ln n ⁡ x d x {\displaystyle \int _{0}^{1}x^{1+\epsilon }(1-x)^{2}\mathrm {d} x=\sum _{n=0}^{\infty }{\frac {\epsilon ^{n}}{n!}}\int _{0}^{1}x(1-x)^{2}\ln ^{n}x\mathrm {d} x} \int _{{0}}^{{1}}x^{{1+\epsilon }}(1-x)^{{2}}{\mathrm {d}}x=\sum _{{n=0}}^{{\infty }}{\frac {\epsilon ^{{n}}}{n!}}\int _{{0}}^{{1}}x(1-x)^{{2}}\ln ^{{n}}x{\mathrm {d}}x ∫ 0 1 x 1 + ϵ ( 1 − x ) 2 d x = Γ ( 2 + ϵ ) Γ ( 3 ) Γ ( 5 + ϵ ) {\displaystyle \int _{0}^{1}x^{1+\epsilon }(1-x)^{2}\mathrm {d} x={\frac {\Gamma (2+\epsilon )\Gamma (3)}{\Gamma (5+\epsilon )}}} \int _{{0}}^{{1}}x^{{1+\epsilon }}(1-x)^{{2}}{\mathrm {d}}x={\frac {\Gamma (2+\epsilon )\Gamma (3)}{\Gamma (5+\epsilon )}} As seen above, we want to find the coefficient of ϵ . {\displaystyle \epsilon .} \epsilon .

Expand the Gamma function into its Taylor series up to first order. Since we are only finding the integral with the log to the first order, we can rewrite the terms in parentheses as exponential functions. Γ ( 2 + ϵ ) Γ ( 3 ) Γ ( 5 + ϵ ) = 2 Γ ( 2 + ϵ ) ( 4 + ϵ ) ( 3 + ϵ ) ( 2 + ϵ ) Γ ( 2 + ϵ ) = 2 4 ⋅ 3 ⋅ 2 1 ( 1 + ϵ / 4 ) ( 1 + ϵ / 3 ) ( 1 + ϵ / 2 ) ≈ 1 12 e − ( 1 4 + 1 3 + 1 2 ) ϵ ≈ 1 12 ( 1 − 13 12 ϵ ) {\displaystyle {\begin{aligned}{\frac {\Gamma (2+\epsilon )\Gamma (3)}{\Gamma (5+\epsilon )}}&={\frac {2\,\Gamma (2+\epsilon )}{(4+\epsilon )(3+\epsilon )(2+\epsilon )\Gamma (2+\epsilon )}}\\&={\frac {2}{4\cdot 3\cdot 2}}{\frac {1}{(1+\epsilon /4)(1+\epsilon /3)(1+\epsilon /2)}}\\&\approx {\frac {1}{12}}e^{-\left({\frac {1}{4}}+{\frac {1}{3}}+{\frac {1}{2}}\right)\epsilon }\\&\approx {\frac {1}{12}}\left(1-{\frac {13}{12}}\epsilon \right)\end{aligned}}} {\begin{aligned}{\frac {\Gamma (2+\epsilon )\Gamma (3)}{\Gamma (5+\epsilon )}}&={\frac {2\,\Gamma (2+\epsilon )}{(4+\epsilon )(3+\epsilon )(2+\epsilon )\Gamma (2+\epsilon )}}\\&={\frac {2}{4\cdot 3\cdot 2}}{\frac {1}{(1+\epsilon /4)(1+\epsilon /3)(1+\epsilon /2)}}\\&\approx {\frac {1}{12}}e^{{-\left({\frac {1}{4}}+{\frac {1}{3}}+{\frac {1}{2}}\right)\epsilon }}\\&\approx {\frac {1}{12}}\left(1-{\frac {13}{12}}\epsilon \right)\end{aligned}}

Evaluate the integral by comparing coefficients. Our answer comes right out of our work. ∫ 0 1 x ( 1 − x ) 2 ln ⁡ x d x = − 13 144 {\displaystyle \int _{0}^{1}x(1-x)^{2}\ln x\mathrm {d} x=-{\frac {13}{144}}} \int _{{0}}^{{1}}x(1-x)^{{2}}\ln x{\mathrm {d}}x=-{\frac {13}{144}} As usual, we get this integral for free, which can be evaluated the standard way. ∫ 0 1 x ( 1 − x ) 2 d x = 1 12 {\displaystyle \int _{0}^{1}x(1-x)^{2}\mathrm {d} x={\frac {1}{12}}} \int _{{0}}^{{1}}x(1-x)^{{2}}{\mathrm {d}}x={\frac {1}{12}}

Deriving Legendre's Duplication Formula

Begin with the integral below. We set z = α = β . {\displaystyle z=\alpha =\beta .} z=\alpha =\beta . ∫ 0 1 t z − 1 ( 1 − t ) z − 1 d t = Γ 2 ( z ) Γ ( 2 z ) {\displaystyle \int _{0}^{1}t^{z-1}(1-t)^{z-1}\mathrm {d} t={\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}} \int _{{0}}^{{1}}t^{{z-1}}(1-t)^{{z-1}}{\mathrm {d}}t={\frac {\Gamma ^{{2}}(z)}{\Gamma (2z)}}

Make the u-sub u = 2 t − 1 {\displaystyle u=2t-1} u=2t-1. Γ 2 ( z ) Γ ( 2 z ) = ∫ 0 1 t z − 1 ( 1 − t ) z − 1 d t = 1 2 2 z − 1 ∫ − 1 1 ( 1 − u 2 ) z − 1 d u = 1 2 2 z − 2 ∫ 0 1 ( 1 − u 2 ) z − 1 d u {\displaystyle {\begin{aligned}{\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}&=\int _{0}^{1}t^{z-1}(1-t)^{z-1}\mathrm {d} t\\&={\frac {1}{2^{2z-1}}}\int _{-1}^{1}(1-u^{2})^{z-1}\mathrm {d} u\\&={\frac {1}{2^{2z-2}}}\int _{0}^{1}(1-u^{2})^{z-1}\mathrm {d} u\end{aligned}}} {\begin{aligned}{\frac {\Gamma ^{{2}}(z)}{\Gamma (2z)}}&=\int _{{0}}^{{1}}t^{{z-1}}(1-t)^{{z-1}}{\mathrm {d}}t\\&={\frac {1}{2^{{2z-1}}}}\int _{{-1}}^{{1}}(1-u^{{2}})^{{z-1}}{\mathrm {d}}u\\&={\frac {1}{2^{{2z-2}}}}\int _{{0}}^{{1}}(1-u^{{2}})^{{z-1}}{\mathrm {d}}u\end{aligned}}

Make a further substitution s = u 2 {\displaystyle s=u^{2}} s=u^{{2}}. Then we can get the integral into the form where we can directly use the Beta function. Γ 2 ( z ) Γ ( 2 z ) = 1 2 2 z − 2 ∫ 0 1 ( 1 − u 2 ) z − 1 d u = 1 2 2 z − 1 ∫ 0 1 s − 1 / 2 ( 1 − s ) z − 1 d s = π 2 2 z − 1 Γ ( z ) Γ ( z + 1 / 2 ) {\displaystyle {\begin{aligned}{\frac {\Gamma ^{2}(z)}{\Gamma (2z)}}&={\frac {1}{2^{2z-2}}}\int _{0}^{1}(1-u^{2})^{z-1}\mathrm {d} u\\&={\frac {1}{2^{2z-1}}}\int _{0}^{1}s^{-1/2}(1-s)^{z-1}\mathrm {d} s\\&={\frac {\sqrt {\pi }}{2^{2z-1}}}{\frac {\Gamma (z)}{\Gamma (z+1/2)}}\end{aligned}}} {\begin{aligned}{\frac {\Gamma ^{{2}}(z)}{\Gamma (2z)}}&={\frac {1}{2^{{2z-2}}}}\int _{{0}}^{{1}}(1-u^{{2}})^{{z-1}}{\mathrm {d}}u\\&={\frac {1}{2^{{2z-1}}}}\int _{{0}}^{{1}}s^{{-1/2}}(1-s)^{{z-1}}{\mathrm {d}}s\\&={\frac {{\sqrt {\pi }}}{2^{{2z-1}}}}{\frac {\Gamma (z)}{\Gamma (z+1/2)}}\end{aligned}} This is Legendre's duplication formula. It allows us to evaluate certain integrals that give us a Γ ( 1 / 2 + ϵ ) {\displaystyle \Gamma (1/2+\epsilon )} \Gamma (1/2+\epsilon ) during our work.

Integrals with Two Logarithms

Evaluate the integral below. We can also use the Beta function to determine integrals like these. ∫ 0 1 ln 2 ⁡ x ln ⁡ ( 1 − x ) d x {\displaystyle \int _{0}^{1}\ln ^{2}x\ln(1-x)\mathrm {d} x} \int _{{0}}^{{1}}\ln ^{{2}}x\ln(1-x){\mathrm {d}}x

Consider the integrals below. Since we have two logs, we need to introduce two parameters. ∫ 0 1 x ϵ ( 1 − x ) δ d x = ∑ m = 0 ∞ ∑ n = 0 ∞ ϵ m m ! δ n n ! ∫ 0 1 ln m ⁡ x ln n ⁡ ( 1 − x ) d x {\displaystyle \int _{0}^{1}x^{\epsilon }(1-x)^{\delta }\mathrm {d} x=\sum _{m=0}^{\infty }\sum _{n=0}^{\infty }{\frac {\epsilon ^{m}}{m!}}{\frac {\delta ^{n}}{n!}}\int _{0}^{1}\ln ^{m}x\ln ^{n}(1-x)\mathrm {d} x} \int _{{0}}^{{1}}x^{{\epsilon }}(1-x)^{{\delta }}{\mathrm {d}}x=\sum _{{m=0}}^{{\infty }}\sum _{{n=0}}^{{\infty }}{\frac {\epsilon ^{{m}}}{m!}}{\frac {\delta ^{{n}}}{n!}}\int _{{0}}^{{1}}\ln ^{{m}}x\ln ^{{n}}(1-x){\mathrm {d}}x ∫ 0 1 x ϵ ( 1 − x ) δ d x = Γ ( 1 + ϵ ) Γ ( 1 + δ ) Γ ( 2 + ϵ + δ ) = Γ ( 1 + ϵ ) Γ ( 1 + δ ) ( 1 + ϵ + δ ) Γ ( 1 + ϵ + δ ) {\displaystyle \int _{0}^{1}x^{\epsilon }(1-x)^{\delta }\mathrm {d} x={\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{\Gamma (2+\epsilon +\delta )}}={\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{(1+\epsilon +\delta )\Gamma (1+\epsilon +\delta )}}} \int _{{0}}^{{1}}x^{{\epsilon }}(1-x)^{{\delta }}{\mathrm {d}}x={\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{\Gamma (2+\epsilon +\delta )}}={\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{(1+\epsilon +\delta )\Gamma (1+\epsilon +\delta )}} Our integral implies that we need to find the coefficient of ϵ 2 δ {\displaystyle \epsilon ^{2}\delta } \epsilon ^{{2}}\delta in the expansion, setting m = 2 {\displaystyle m=2} m=2 and n = 1. {\displaystyle n=1.} n=1. Furthermore, we must multiply the eventual result we obtain by the factorial of the power. In this case, 2 ! 1 ! = 2. {\displaystyle 2!1!=2.} 2!1!=2.

Expand the Gamma functions and the fraction. We see that the terms including the Euler-Mascheroni constant vanish. Furthermore, the terms in the sum cancel in a way such that only the cross terms are left intact. (We break up the exponential function into two to save space.) The fraction is expanded into its power series. Γ ( 1 + ϵ ) Γ ( 1 + δ ) Γ ( 1 + ϵ + δ ) = exp ⁡ [ − γ δ − γ ϵ + ∑ k = 2 ∞ ( − 1 ) k ζ ( k ) k ( δ k + ϵ k ) ] ∗ exp ⁡ [ γ ( δ + ϵ ) − ∑ k = 2 ∞ ( − 1 ) k ζ ( k ) k ( δ + ϵ ) k ] = exp ⁡ [ ∑ k = 2 ∞ ( − 1 ) k ζ ( k ) k ( δ k + ϵ k − ( δ + ϵ ) k ) ] {\displaystyle {\begin{aligned}{\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{\Gamma (1+\epsilon +\delta )}}&=\exp \left[-\gamma \delta -\gamma \epsilon +\sum _{k=2}^{\infty }{\frac {(-1)^{k}\zeta (k)}{k}}(\delta ^{k}+\epsilon ^{k})\right]\\&\ *\,\exp \left[\gamma (\delta +\epsilon )-\sum _{k=2}^{\infty }{\frac {(-1)^{k}\zeta (k)}{k}}(\delta +\epsilon )^{k}\right]\\&=\exp \left[\sum _{k=2}^{\infty }{\frac {(-1)^{k}\zeta (k)}{k}}(\delta ^{k}+\epsilon ^{k}-(\delta +\epsilon )^{k})\right]\end{aligned}}} {\begin{aligned}{\frac {\Gamma (1+\epsilon )\Gamma (1+\delta )}{\Gamma (1+\epsilon +\delta )}}&=\exp \left[-\gamma \delta -\gamma \epsilon +\sum _{{k=2}}^{{\infty }}{\frac {(-1)^{{k}}\zeta (k)}{k}}(\delta ^{{k}}+\epsilon ^{{k}})\right]\\&\ *\,\exp \left[\gamma (\delta +\epsilon )-\sum _{{k=2}}^{{\infty }}{\frac {(-1)^{{k}}\zeta (k)}{k}}(\delta +\epsilon )^{{k}}\right]\\&=\exp \left[\sum _{{k=2}}^{{\infty }}{\frac {(-1)^{{k}}\zeta (k)}{k}}(\delta ^{{k}}+\epsilon ^{{k}}-(\delta +\epsilon )^{{k}})\right]\end{aligned}} 1 1 + ϵ + δ = 1 − ( ϵ + δ ) + ( ϵ + δ ) 2 − ( ϵ + δ ) 3 + ⋯ {\displaystyle {\frac {1}{1+\epsilon +\delta }}=1-(\epsilon +\delta )+(\epsilon +\delta )^{2}-(\epsilon +\delta )^{3}+\cdots } {\frac {1}{1+\epsilon +\delta }}=1-(\epsilon +\delta )+(\epsilon +\delta )^{{2}}-(\epsilon +\delta )^{{3}}+\cdots

Add the coefficients of ϵ 2 δ {\displaystyle \epsilon ^{2}\delta } \epsilon ^{{2}}\delta . We need only terms up to k = 3 , {\displaystyle k=3,} k=3, and the Taylor series of that exponential function only goes up to first-order. We will also need terms of the power series up to third order. Remember that we do not need to multiply everything out. We are only interested in the coefficients of ϵ 2 δ . {\displaystyle \epsilon ^{2}\delta .} \epsilon ^{{2}}\delta . Be sure to keep track of the signs. ϵ 2 δ : ζ ( 3 ) + ζ ( 2 ) − 3 {\displaystyle \epsilon ^{2}\delta :\zeta (3)+\zeta (2)-3} \epsilon ^{{2}}\delta :\zeta (3)+\zeta (2)-3 Remembering to multiply by 2 to account for the factorial on ϵ 2 , {\displaystyle \epsilon ^{2},} \epsilon ^{{2}}, this immediately gets us the desired result. ∫ 0 1 ln 2 ⁡ x ln ⁡ ( 1 − x ) d x = 2 ( ζ ( 3 ) + ζ ( 2 ) − 3 ) {\displaystyle \int _{0}^{1}\ln ^{2}x\ln(1-x)\mathrm {d} x=2(\zeta (3)+\zeta (2)-3)} \int _{{0}}^{{1}}\ln ^{{2}}x\ln(1-x){\mathrm {d}}x=2(\zeta (3)+\zeta (2)-3)

Verify the integrals below. We can also show similar integrals using this technique. For the first one, we find coefficients of ϵ δ . {\displaystyle \epsilon \delta .} \epsilon \delta . For the second one, we find coefficients of ϵ 2 δ 2 . {\displaystyle \epsilon ^{2}\delta ^{2}.} \epsilon ^{{2}}\delta ^{{2}}. In principle, it is possible to evaluate integrals like these with any integer power on the logs. We would just have to keep more terms in our evaluation. ∫ 0 1 ln ⁡ x ln ⁡ ( 1 − x ) d x = 2 − ζ ( 2 ) {\displaystyle \int _{0}^{1}\ln x\ln(1-x)\mathrm {d} x=2-\zeta (2)} \int _{{0}}^{{1}}\ln x\ln(1-x){\mathrm {d}}x=2-\zeta (2) ∫ 0 1 ln 2 ⁡ x ln 2 ⁡ ( 1 − x ) d x = 24 − 8 ζ ( 2 ) − 8 ζ ( 3 ) + 2 ζ 2 ( 2 ) − 6 ζ ( 4 ) {\displaystyle \int _{0}^{1}\ln ^{2}x\ln ^{2}(1-x)\mathrm {d} x=24-8\zeta (2)-8\zeta (3)+2\zeta ^{2}(2)-6\zeta (4)} \int _{{0}}^{{1}}\ln ^{{2}}x\ln ^{{2}}(1-x){\mathrm {d}}x=24-8\zeta (2)-8\zeta (3)+2\zeta ^{{2}}(2)-6\zeta (4)

Change of Variables

Begin with the Beta function integral. In this section, we will show a u-sub that converts the Beta function into an integral from 0 to infinity, which will produce some very interesting results. ∫ 0 1 t α − 1 ( 1 − t ) β − 1 d t {\displaystyle \int _{0}^{1}t^{\alpha -1}(1-t)^{\beta -1}\mathrm {d} t} \int _{{0}}^{{1}}t^{{\alpha -1}}(1-t)^{{\beta -1}}{\mathrm {d}}t

Make the u-sub u = 1 − t t {\displaystyle u={\frac {1-t}{t}}} u={\frac {1-t}{t}}. This does two things. First, it allows us to directly evaluate integrals with 1 + u {\displaystyle 1+u} 1+u in the denominator, which was previously not allowed. Second, it changes the boundaries. The way we now evaluate is to find β {\displaystyle \beta } \beta first, and then find α , {\displaystyle \alpha ,} \alpha , because of this substitution. Γ ( α ) Γ ( β ) Γ ( α + β ) = ∫ 0 1 t α − 1 ( 1 − t ) β − 1 d t = ∫ 0 ∞ 1 ( 1 + u ) α + 1 ( 1 − 1 1 + u ) β − 1 d u = ∫ 0 ∞ u β − 1 ( 1 + u ) α + β d u {\displaystyle {\begin{aligned}{\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}&=\int _{0}^{1}t^{\alpha -1}(1-t)^{\beta -1}\mathrm {d} t\\&=\int _{0}^{\infty }{\frac {1}{(1+u)^{\alpha +1}}}\left(1-{\frac {1}{1+u}}\right)^{\beta -1}\mathrm {d} u\\&=\int _{0}^{\infty }{\frac {u^{\beta -1}}{(1+u)^{\alpha +\beta }}}\mathrm {d} u\end{aligned}}} {\begin{aligned}{\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}&=\int _{{0}}^{{1}}t^{{\alpha -1}}(1-t)^{{\beta -1}}{\mathrm {d}}t\\&=\int _{{0}}^{{\infty }}{\frac {1}{(1+u)^{{\alpha +1}}}}\left(1-{\frac {1}{1+u}}\right)^{{\beta -1}}{\mathrm {d}}u\\&=\int _{{0}}^{{\infty }}{\frac {u^{{\beta -1}}}{(1+u)^{{\alpha +\beta }}}}{\mathrm {d}}u\end{aligned}}

Verify the integrals below. This form of the Beta function allows direct access to another class of integrals otherwise only accessible via residues. We can use Euler's reflection formula to simplify integrals, particularly the second one listed. ∫ 0 ∞ 1 x ( 1 + x ) 3 d x = 2 {\displaystyle \int _{0}^{\infty }{\frac {1}{\sqrt {x(1+x)^{3}}}}\mathrm {d} x=2} \int _{{0}}^{{\infty }}{\frac {1}{{\sqrt {x(1+x)^{{3}}}}}}{\mathrm {d}}x=2 ∫ 0 ∞ x 2 x 4 ( x + 1 ) 4 d x = 15 2 π 128 {\displaystyle \int _{0}^{\infty }{\frac {x^{2}{\sqrt[{4}]{x}}}{(x+1)^{4}}}\mathrm {d} x={\frac {15{\sqrt {2}}\pi }{128}}} \int _{{0}}^{{\infty }}{\frac {x^{{2}}{\sqrt[ {4}]{x}}}{(x+1)^{{4}}}}{\mathrm {d}}x={\frac {15{\sqrt {2}}\pi }{128}}

Consider the integral below. We replace the term in the denominator with x n + 1 , {\displaystyle x^{n}+1,} x^{{n}}+1, which after a u-sub, leads to more general results, since we can differentiate under the integral with respect to any of the three parameters. In particular, when we set α = 1 , {\displaystyle \alpha =1,} \alpha =1, we arrive at a very attractive answer involving the cosecant function (which we use the reflection formula to derive). ∫ 0 ∞ x m − 1 ( x n + 1 ) α d x = Γ ( m / n ) Γ ( α − m / n ) n Γ ( α ) {\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}}{(x^{n}+1)^{\alpha }}}\mathrm {d} x={\frac {\Gamma (m/n)\Gamma (\alpha -m/n)}{n\Gamma (\alpha )}}} \int _{{0}}^{{\infty }}{\frac {x^{{m-1}}}{(x^{{n}}+1)^{{\alpha }}}}{\mathrm {d}}x={\frac {\Gamma (m/n)\Gamma (\alpha -m/n)}{n\Gamma (\alpha )}} ∫ 0 ∞ x m − 1 x n + 1 d x = π n csc ⁡ m π n {\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}}{x^{n}+1}}\mathrm {d} x={\frac {\pi }{n}}\csc {\frac {m\pi }{n}}} \int _{{0}}^{{\infty }}{\frac {x^{{m-1}}}{x^{{n}}+1}}{\mathrm {d}}x={\frac {\pi }{n}}\csc {\frac {m\pi }{n}} These results can directly be used to evaluate more integrals. Verify these. ∫ 0 ∞ x 2 x ( x 3 + 1 ) 2 d x = π 9 {\displaystyle \int _{0}^{\infty }{\frac {x^{2}{\sqrt {x}}}{(x^{3}+1)^{2}}}\mathrm {d} x={\frac {\pi }{9}}} \int _{{0}}^{{\infty }}{\frac {x^{{2}}{\sqrt {x}}}{(x^{{3}}+1)^{{2}}}}{\mathrm {d}}x={\frac {\pi }{9}} ∫ 0 ∞ 1 x 4 + 1 d x = π 2 2 {\displaystyle \int _{0}^{\infty }{\frac {1}{x^{4}+1}}\mathrm {d} x={\frac {\pi }{2{\sqrt {2}}}}} \int _{{0}}^{{\infty }}{\frac {1}{x^{{4}}+1}}{\mathrm {d}}x={\frac {\pi }{2{\sqrt {2}}}}

Differentiate under the integral with respect to m {\displaystyle m} m. The above result with the cosecant is a very potent integral because we can also differentiate once and twice to obtain some more results involving logs. (We use a trig identity to simplify the result after differentiating twice.) ∫ 0 ∞ x m − 1 ln ⁡ x x n + 1 d x = − π 2 n 2 csc ⁡ m π n cot ⁡ m π n {\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}\ln x}{x^{n}+1}}\mathrm {d} x=-{\frac {\pi ^{2}}{n^{2}}}\csc {\frac {m\pi }{n}}\cot {\frac {m\pi }{n}}} \int _{{0}}^{{\infty }}{\frac {x^{{m-1}}\ln x}{x^{{n}}+1}}{\mathrm {d}}x=-{\frac {\pi ^{{2}}}{n^{{2}}}}\csc {\frac {m\pi }{n}}\cot {\frac {m\pi }{n}} ∫ 0 ∞ x m − 1 ln 2 ⁡ x x n + 1 d x = π 3 n 3 csc ⁡ m π n ( 2 csc 2 ⁡ m π n − 1 ) {\displaystyle \int _{0}^{\infty }{\frac {x^{m-1}\ln ^{2}x}{x^{n}+1}}\mathrm {d} x={\frac {\pi ^{3}}{n^{3}}}\csc {\frac {m\pi }{n}}\left(2\csc ^{2}{\frac {m\pi }{n}}-1\right)} \int _{{0}}^{{\infty }}{\frac {x^{{m-1}}\ln ^{{2}}x}{x^{{n}}+1}}{\mathrm {d}}x={\frac {\pi ^{{3}}}{n^{{3}}}}\csc {\frac {m\pi }{n}}\left(2\csc ^{{2}}{\frac {m\pi }{n}}-1\right) Use these results to verify the integrals below. These integrals have extremely complicated antiderivatives, and there is virtually no hope of approaching them from the perspective of the fundamental theorem. However, these extremely simple answers only showcases the power of the Beta function - it makes the process of obtaining a simple answer, simple. ∫ 0 ∞ x 2 ln ⁡ x x 4 + 1 d x = π 2 2 16 {\displaystyle \int _{0}^{\infty }{\frac {x^{2}\ln x}{x^{4}+1}}\mathrm {d} x={\frac {\pi ^{2}{\sqrt {2}}}{16}}} \int _{{0}}^{{\infty }}{\frac {x^{{2}}\ln x}{x^{{4}}+1}}{\mathrm {d}}x={\frac {\pi ^{{2}}{\sqrt {2}}}{16}} ∫ 0 ∞ ln 2 ⁡ x x 3 + 1 d x = 10 π 3 81 3 {\displaystyle \int _{0}^{\infty }{\frac {\ln ^{2}x}{x^{3}+1}}\mathrm {d} x={\frac {10\pi ^{3}}{81{\sqrt {3}}}}} \int _{{0}}^{{\infty }}{\frac {\ln ^{{2}}x}{x^{{3}}+1}}{\mathrm {d}}x={\frac {10\pi ^{{3}}}{81{\sqrt {3}}}} ∫ 0 ∞ x 3 ln 2 ⁡ x ( x 4 + 1 ) 2 d x = π 2 192 {\displaystyle \int _{0}^{\infty }{\frac {x^{3}\ln ^{2}x}{(x^{4}+1)^{2}}}\mathrm {d} x={\frac {\pi ^{2}}{192}}} \int _{{0}}^{{\infty }}{\frac {x^{{3}}\ln ^{{2}}x}{(x^{{4}}+1)^{{2}}}}{\mathrm {d}}x={\frac {\pi ^{{2}}}{192}}

Trigonometric Form

Begin with the product of two Gamma functions. If you are familiar with the derivation of the Beta function, we start from the same place. However, we switch to polar and make a substitution to get a trigonometric integral. Γ ( α ) Γ ( β ) = ∫ 0 ∞ u α − 1 e − u d u ∫ 0 ∞ v β − 1 e − v d v {\displaystyle \Gamma (\alpha )\Gamma (\beta )=\int _{0}^{\infty }u^{\alpha -1}e^{-u}\mathrm {d} u\int _{0}^{\infty }v^{\beta -1}e^{-v}\mathrm {d} v} \Gamma (\alpha )\Gamma (\beta )=\int _{{0}}^{{\infty }}u^{{\alpha -1}}e^{{-u}}{\mathrm {d}}u\int _{{0}}^{{\infty }}v^{{\beta -1}}e^{{-v}}{\mathrm {d}}v

Make the u-subs u = x 2 {\displaystyle u=x^{2}} u=x^{{2}} and v = y 2 {\displaystyle v=y^{2}} v=y^{{2}} and switch to polar. Recall that the area element d x d y = r d r d θ {\displaystyle \mathrm {d} x\mathrm {d} y=r\mathrm {d} r\mathrm {d} \theta } {\mathrm {d}}x{\mathrm {d}}y=r{\mathrm {d}}r{\mathrm {d}}\theta and the bounds for θ {\displaystyle \theta } \theta are from 0 {\displaystyle 0} {\displaystyle 0} to π / 2 {\displaystyle \pi /2} \pi /2 because we are integrating over only quadrant I. Γ ( α ) Γ ( β ) = 4 ∫ 0 ∞ x 2 α − 1 e − x 2 d x ∫ 0 ∞ y 2 β − 1 e − y 2 d y = 4 ∫ 0 ∞ r d r r 2 α + 2 β − 2 e − r 2 ∫ 0 π / 2 cos 2 α − 1 ⁡ θ sin 2 β − 1 ⁡ θ d θ {\displaystyle {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=4\int _{0}^{\infty }x^{2\alpha -1}e^{-x^{2}}\mathrm {d} x\int _{0}^{\infty }y^{2\beta -1}e^{-y^{2}}\mathrm {d} y\\&=4\int _{0}^{\infty }r\mathrm {d} r\,r^{2\alpha +2\beta -2}e^{-r^{2}}\int _{0}^{\pi /2}\cos ^{2\alpha -1}\theta \sin ^{2\beta -1}\theta \mathrm {d} \theta \end{aligned}}} {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=4\int _{{0}}^{{\infty }}x^{{2\alpha -1}}e^{{-x^{{2}}}}{\mathrm {d}}x\int _{{0}}^{{\infty }}y^{{2\beta -1}}e^{{-y^{{2}}}}{\mathrm {d}}y\\&=4\int _{{0}}^{{\infty }}r{\mathrm {d}}r\,r^{{2\alpha +2\beta -2}}e^{{-r^{{2}}}}\int _{{0}}^{{\pi /2}}\cos ^{{2\alpha -1}}\theta \sin ^{{2\beta -1}}\theta {\mathrm {d}}\theta \end{aligned}}

Make the u-sub u = r 2 {\displaystyle u=r^{2}} u=r^{{2}}. After substituting and simplifying, we obtain our desired result. Be careful of the extra 1 / 2. {\displaystyle 1/2.} 1/2. Γ ( α ) Γ ( β ) = 2 ∫ 0 ∞ u α + β − 1 e − u d u ∫ 0 π / 2 cos 2 α − 1 ⁡ θ sin 2 β − 1 ⁡ θ d θ = 2 Γ ( α + β ) ∫ 0 π / 2 cos 2 α − 1 ⁡ θ sin 2 β − 1 ⁡ θ d θ {\displaystyle {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=2\int _{0}^{\infty }u^{\alpha +\beta -1}e^{-u}\mathrm {d} u\int _{0}^{\pi /2}\cos ^{2\alpha -1}\theta \sin ^{2\beta -1}\theta \mathrm {d} \theta \\&=2\,\Gamma (\alpha +\beta )\int _{0}^{\pi /2}\cos ^{2\alpha -1}\theta \sin ^{2\beta -1}\theta \mathrm {d} \theta \end{aligned}}} {\begin{aligned}\Gamma (\alpha )\Gamma (\beta )&=2\int _{{0}}^{{\infty }}u^{{\alpha +\beta -1}}e^{{-u}}{\mathrm {d}}u\int _{{0}}^{{\pi /2}}\cos ^{{2\alpha -1}}\theta \sin ^{{2\beta -1}}\theta {\mathrm {d}}\theta \\&=2\,\Gamma (\alpha +\beta )\int _{{0}}^{{\pi /2}}\cos ^{{2\alpha -1}}\theta \sin ^{{2\beta -1}}\theta {\mathrm {d}}\theta \end{aligned}} Γ ( α ) Γ ( β ) 2 Γ ( α + β ) = ∫ 0 π / 2 cos 2 α − 1 ⁡ θ sin 2 β − 1 ⁡ θ d θ {\displaystyle {\frac {\Gamma (\alpha )\Gamma (\beta )}{2\,\Gamma (\alpha +\beta )}}=\int _{0}^{\pi /2}\cos ^{2\alpha -1}\theta \sin ^{2\beta -1}\theta \mathrm {d} \theta } {\frac {\Gamma (\alpha )\Gamma (\beta )}{2\,\Gamma (\alpha +\beta )}}=\int _{{0}}^{{\pi /2}}\cos ^{{2\alpha -1}}\theta \sin ^{{2\beta -1}}\theta {\mathrm {d}}\theta This is a very important result, and one that is very often used with integer powers, which provide very "nice" answers.

Verify the following integrals. These are daunting with reduction of power formulae and other techniques, but trivial from the perspective of the Beta function. ∫ 0 π / 2 cos 4 ⁡ x sin 5 ⁡ x d x = 8 315 {\displaystyle \int _{0}^{\pi /2}\cos ^{4}x\sin ^{5}x\mathrm {d} x={\frac {8}{315}}} \int _{{0}}^{{\pi /2}}\cos ^{{4}}x\sin ^{{5}}x{\mathrm {d}}x={\frac {8}{315}} ∫ 0 π / 2 sin ⁡ x d x = 2 π Γ 2 ( 3 / 4 ) {\displaystyle \int _{0}^{\pi /2}{\sqrt {\sin x}}\mathrm {d} x={\sqrt {\frac {2}{\pi }}}\,\Gamma ^{2}(3/4)} \int _{{0}}^{{\pi /2}}{\sqrt {\sin x}}{\mathrm {d}}x={\sqrt {{\frac {2}{\pi }}}}\,\Gamma ^{{2}}(3/4) ∫ 0 π / 2 sin 4 ⁡ x cos 2 ⁡ x sin ⁡ x d x = 28 Γ 2 ( 3 / 4 ) 195 2 π {\displaystyle \int _{0}^{\pi /2}\sin ^{4}x\cos ^{2}x{\sqrt {\sin x}}\mathrm {d} x={\frac {28\,\Gamma ^{2}(3/4)}{195{\sqrt {2\pi }}}}} \int _{{0}}^{{\pi /2}}\sin ^{{4}}x\cos ^{{2}}x{\sqrt {\sin x}}{\mathrm {d}}x={\frac {28\,\Gamma ^{{2}}(3/4)}{195{\sqrt {2\pi }}}}

Logarithms of Trigonometric Functions

Evaluate the integral below. The integral contains a composition of functions whose antiderivative cannot be written in terms of elementary functions. Nevertheless, the integral contains an exact solution. ∫ 0 π / 2 ln ⁡ sin ⁡ x d x {\displaystyle \int _{0}^{\pi /2}\ln \sin x\mathrm {d} x} \int _{{0}}^{{\pi /2}}\ln \sin x{\mathrm {d}}x

Consider the integrals below. As usual, we begin with the more general case of expanding into a series, neglecting higher-order terms, and finding the appropriate coefficient. These integrals will require the use of the duplication formula. ∫ 0 π / 2 sin ϵ ⁡ x d x = ∑ n = 0 ∞ ϵ n n ! ∫ 0 π / 2 ln n ⁡ sin ⁡ x d x {\displaystyle \int _{0}^{\pi /2}\sin ^{\epsilon }x\mathrm {d} x=\sum _{n=0}^{\infty }{\frac {\epsilon ^{n}}{n!}}\int _{0}^{\pi /2}\ln ^{n}\sin x\mathrm {d} x} \int _{{0}}^{{\pi /2}}\sin ^{{\epsilon }}x{\mathrm {d}}x=\sum _{{n=0}}^{{\infty }}{\frac {\epsilon ^{{n}}}{n!}}\int _{{0}}^{{\pi /2}}\ln ^{{n}}\sin x{\mathrm {d}}x ∫ 0 π / 2 sin ϵ ⁡ x d x = Γ ( 1 / 2 ) Γ ( 1 / 2 + ϵ / 2 ) 2 Γ ( 1 + ϵ / 2 ) {\displaystyle \int _{0}^{\pi /2}\sin ^{\epsilon }x\mathrm {d} x={\frac {\Gamma (1/2)\Gamma (1/2+\epsilon /2)}{2\,\Gamma (1+\epsilon /2)}}} \int _{{0}}^{{\pi /2}}\sin ^{{\epsilon }}x{\mathrm {d}}x={\frac {\Gamma (1/2)\Gamma (1/2+\epsilon /2)}{2\,\Gamma (1+\epsilon /2)}}

Expand to the first order. After using the duplication formula, we see that the ratio Γ ( 1 + ϵ ) Γ 2 ( 1 + ϵ / 2 ) {\displaystyle {\frac {\Gamma (1+\epsilon )}{\Gamma ^{2}(1+\epsilon /2)}}} {\frac {\Gamma (1+\epsilon )}{\Gamma ^{{2}}(1+\epsilon /2)}} cancels up to the first order, leaving us with a very simple expansion. Γ ( 1 / 2 ) Γ ( 1 / 2 + ϵ / 2 ) 2 Γ ( 1 + ϵ / 2 ) = π 2 ⋅ 2 ϵ Γ ( 1 + ϵ ) Γ 2 ( 1 + ϵ / 2 ) ≈ π 2 e − ϵ ln ⁡ 2 ≈ π 2 ( 1 − ϵ ln ⁡ 2 ) {\displaystyle {\begin{aligned}{\frac {\Gamma (1/2)\Gamma (1/2+\epsilon /2)}{2\,\Gamma (1+\epsilon /2)}}&={\frac {\pi }{2\cdot 2^{\epsilon }}}{\frac {\Gamma (1+\epsilon )}{\Gamma ^{2}(1+\epsilon /2)}}\\&\approx {\frac {\pi }{2}}e^{-\epsilon \ln 2}\\&\approx {\frac {\pi }{2}}(1-\epsilon \ln 2)\end{aligned}}} {\begin{aligned}{\frac {\Gamma (1/2)\Gamma (1/2+\epsilon /2)}{2\,\Gamma (1+\epsilon /2)}}&={\frac {\pi }{2\cdot 2^{{\epsilon }}}}{\frac {\Gamma (1+\epsilon )}{\Gamma ^{{2}}(1+\epsilon /2)}}\\&\approx {\frac {\pi }{2}}e^{{-\epsilon \ln 2}}\\&\approx {\frac {\pi }{2}}(1-\epsilon \ln 2)\end{aligned}}

Evaluate by equating coefficients. ∫ 0 π / 2 ln ⁡ sin ⁡ x d x = − π 2 ln ⁡ 2 {\displaystyle \int _{0}^{\pi /2}\ln \sin x\mathrm {d} x=-{\frac {\pi }{2}}\ln 2} \int _{{0}}^{{\pi /2}}\ln \sin x{\mathrm {d}}x=-{\frac {\pi }{2}}\ln 2

Verify the following integrals. This technique can once again be used to evaluate the entire class of integrals. ∫ 0 π / 2 sin ⁡ x ln ⁡ sin ⁡ x d x = ln ⁡ 2 − 1 {\displaystyle \int _{0}^{\pi /2}\sin x\ln \sin x\mathrm {d} x=\ln 2-1} \int _{{0}}^{{\pi /2}}\sin x\ln \sin x{\mathrm {d}}x=\ln 2-1 ∫ 0 π / 2 sin 2 ⁡ x cos 2 ⁡ x ln ⁡ sin ⁡ x d x = π 64 ( 1 − 4 ln ⁡ 2 ) {\displaystyle \int _{0}^{\pi /2}\sin ^{2}x\cos ^{2}x\ln \sin x\mathrm {d} x={\frac {\pi }{64}}(1-4\ln 2)} \int _{{0}}^{{\pi /2}}\sin ^{{2}}x\cos ^{{2}}x\ln \sin x{\mathrm {d}}x={\frac {\pi }{64}}(1-4\ln 2)

Regularization

Evaluate the integral below. This is an example of an integral that converges, but we cannot directly apply our techniques to evaluate because the integral that we would've considered does not converge. ∫ 0 π / 2 ln ⁡ sin ⁡ x cos ⁡ x d x {\displaystyle \int _{0}^{\pi /2}{\frac {\ln \sin x}{\cos x}}\mathrm {d} x} \int _{{0}}^{{\pi /2}}{\frac {\ln \sin x}{\cos x}}{\mathrm {d}}x

Consider the regularized integral. We need to add a term δ {\displaystyle \delta } \delta that "tames" the integral so that it converges. Otherwise, we would get a Γ ( 0 ) {\displaystyle \Gamma (0)} \Gamma (0) term that is undefined. Here, δ {\displaystyle \delta } \delta is a small number that is taken to be 0 at a convenient time. ∫ 0 π / 2 sin ϵ ⁡ x cos − 1 + δ ⁡ x d x = Γ ( 1 + ϵ 2 ) Γ ( δ 2 ) 2 Γ ( 1 2 + ϵ + δ 2 ) {\displaystyle \int _{0}^{\pi /2}\sin ^{\epsilon }x\cos ^{-1+\delta }x\mathrm {d} x={\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left({\frac {\delta }{2}}\right)}{2\,\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}} \int _{{0}}^{{\pi /2}}\sin ^{{\epsilon }}x\cos ^{{-1+\delta }}x{\mathrm {d}}x={\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left({\frac {\delta }{2}}\right)}{2\,\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}

Multiply the top and bottom by δ / 2 {\displaystyle \delta /2} \delta /2. This gets our result into a form so that we can use a series expansion around Γ ( 1 ) . {\displaystyle \Gamma (1).} \Gamma (1). Then we use the duplication formula. Γ ( 1 + ϵ 2 ) Γ ( δ 2 ) 2 Γ ( 1 2 + ϵ + δ 2 ) = 1 δ Γ ( 1 + ϵ 2 ) Γ ( 1 + δ 2 ) Γ ( 1 2 + ϵ + δ 2 ) = Γ ( 1 + δ / 2 ) δ π Γ ( 1 + ϵ ) 2 ϵ Γ ( 1 + ϵ / 2 ) π Γ ( 1 + ϵ + δ ) 2 ϵ + δ Γ ( 1 + ϵ + δ 2 ) = 2 δ δ Γ ( 1 + δ / 2 ) Γ ( 1 + ϵ ) Γ ( 1 + ϵ + δ 2 ) Γ ( 1 + ϵ / 2 ) Γ ( 1 + ϵ + δ ) {\displaystyle {\begin{aligned}{\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left({\frac {\delta }{2}}\right)}{2\,\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}&={\frac {1}{\delta }}{\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left(1+{\frac {\delta }{2}}\right)}{\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}\\&={\frac {\Gamma (1+\delta /2)}{\delta }}{\frac {\frac {{\sqrt {\pi }}\,\Gamma (1+\epsilon )}{2^{\epsilon }\Gamma (1+\epsilon /2)}}{\frac {{\sqrt {\pi }}\,\Gamma (1+\epsilon +\delta )}{2^{\epsilon +\delta }\Gamma \left(1+{\frac {\epsilon +\delta }{2}}\right)}}}\\&={\frac {2^{\delta }}{\delta }}{\frac {\Gamma (1+\delta /2)\Gamma (1+\epsilon )\Gamma \left(1+{\frac {\epsilon +\delta }{2}}\right)}{\Gamma (1+\epsilon /2)\Gamma (1+\epsilon +\delta )}}\end{aligned}}} {\begin{aligned}{\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left({\frac {\delta }{2}}\right)}{2\,\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}&={\frac {1}{\delta }}{\frac {\Gamma \left({\frac {1+\epsilon }{2}}\right)\Gamma \left(1+{\frac {\delta }{2}}\right)}{\Gamma \left({\frac {1}{2}}+{\frac {\epsilon +\delta }{2}}\right)}}\\&={\frac {\Gamma (1+\delta /2)}{\delta }}{\frac {{\frac {{\sqrt {\pi }}\,\Gamma (1+\epsilon )}{2^{{\epsilon }}\Gamma (1+\epsilon /2)}}}{{\frac {{\sqrt {\pi }}\,\Gamma (1+\epsilon +\delta )}{2^{{\epsilon +\delta }}\Gamma \left(1+{\frac {\epsilon +\delta }{2}}\right)}}}}\\&={\frac {2^{{\delta }}}{\delta }}{\frac {\Gamma (1+\delta /2)\Gamma (1+\epsilon )\Gamma \left(1+{\frac {\epsilon +\delta }{2}}\right)}{\Gamma (1+\epsilon /2)\Gamma (1+\epsilon +\delta )}}\end{aligned}}

Expand and look for coefficients of ϵ δ {\displaystyle \epsilon \delta } \epsilon \delta . We are interested in the coefficient of ϵ , {\displaystyle \epsilon ,} \epsilon , but we need to find the coefficient of ϵ δ {\displaystyle \epsilon \delta } \epsilon \delta here in order to cancel the δ {\displaystyle \delta } \delta in front. Notice that any higher-order δ {\displaystyle \delta } \delta terms will vanish. ⋯ = e δ ln ⁡ 2 δ exp ⁡ [ ∑ k = 2 ∞ ( − 1 ) k ζ ( k ) k ( ϵ k + ( ϵ + δ 2 ) k + ( δ 2 ) k − ( ϵ 2 ) k − ( ϵ + δ ) k ) ] ≈ 1 δ ( 1 + δ ln ⁡ 2 ) ( 1 + ζ ( 2 ) 2 ( ϵ 2 + ( ϵ + δ 2 ) 2 + ( δ 2 ) 2 − ( ϵ 2 ) 2 − ( ϵ + δ ) 2 ) ) {\displaystyle {\begin{aligned}\cdots &={\frac {e^{\delta \ln 2}}{\delta }}\exp \left[\sum _{k=2}^{\infty }{\frac {(-1)^{k}\zeta (k)}{k}}\left(\epsilon ^{k}+\left({\frac {\epsilon +\delta }{2}}\right)^{k}+\left({\frac {\delta }{2}}\right)^{k}-\left({\frac {\epsilon }{2}}\right)^{k}-(\epsilon +\delta )^{k}\right)\right]\\&\approx {\frac {1}{\delta }}(1+\delta \ln 2)\left(1+{\frac {\zeta (2)}{2}}\left(\epsilon ^{2}+\left({\frac {\epsilon +\delta }{2}}\right)^{2}+\left({\frac {\delta }{2}}\right)^{2}-\left({\frac {\epsilon }{2}}\right)^{2}-(\epsilon +\delta )^{2}\right)\right)\end{aligned}}} {\begin{aligned}\cdots &={\frac {e^{{\delta \ln 2}}}{\delta }}\exp \left[\sum _{{k=2}}^{{\infty }}{\frac {(-1)^{{k}}\zeta (k)}{k}}\left(\epsilon ^{{k}}+\left({\frac {\epsilon +\delta }{2}}\right)^{{k}}+\left({\frac {\delta }{2}}\right)^{{k}}-\left({\frac {\epsilon }{2}}\right)^{{k}}-(\epsilon +\delta )^{{k}}\right)\right]\\&\approx {\frac {1}{\delta }}(1+\delta \ln 2)\left(1+{\frac {\zeta (2)}{2}}\left(\epsilon ^{{2}}+\left({\frac {\epsilon +\delta }{2}}\right)^{{2}}+\left({\frac {\delta }{2}}\right)^{{2}}-\left({\frac {\epsilon }{2}}\right)^{{2}}-(\epsilon +\delta )^{{2}}\right)\right)\end{aligned}} Notice that the δ ln ⁡ 2 {\displaystyle \delta \ln 2} \delta \ln 2 term cannot contribute to the coefficient because there is no ϵ {\displaystyle \epsilon } \epsilon term on the right. Therefore, the only terms that contribute are the cross terms. ζ ( 2 ) 2 ( 1 2 − 2 ) ϵ δ = − 3 4 ζ ( 2 ) ϵ δ {\displaystyle {\frac {\zeta (2)}{2}}\left({\frac {1}{2}}-2\right)\epsilon \delta =-{\frac {3}{4}}\zeta (2)\epsilon \delta } {\frac {\zeta (2)}{2}}\left({\frac {1}{2}}-2\right)\epsilon \delta =-{\frac {3}{4}}\zeta (2)\epsilon \delta

Evaluate by equating coefficients. We can write our answer in terms of π {\displaystyle \pi } \pi by making use of ζ ( 2 ) = π 2 6 . {\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}.} \zeta (2)={\frac {\pi ^{{2}}}{6}}. ∫ 0 π / 2 ln ⁡ sin ⁡ x cos ⁡ x d x = − π 2 8 {\displaystyle \int _{0}^{\pi /2}{\frac {\ln \sin x}{\cos x}}\mathrm {d} x=-{\frac {\pi ^{2}}{8}}} \int _{{0}}^{{\pi /2}}{\frac {\ln \sin x}{\cos x}}{\mathrm {d}}x=-{\frac {\pi ^{{2}}}{8}}

Verify the integral below. The work that was done to evaluate the first integral can be recycled to evaluate this similar integral. ∫ 0 π / 2 ln 2 ⁡ sin ⁡ x cos ⁡ x d x = 7 4 ζ ( 3 ) {\displaystyle \int _{0}^{\pi /2}{\frac {\ln ^{2}\sin x}{\cos x}}\mathrm {d} x={\frac {7}{4}}\zeta (3)} \int _{{0}}^{{\pi /2}}{\frac {\ln ^{{2}}\sin x}{\cos x}}{\mathrm {d}}x={\frac {7}{4}}\zeta (3)