Evaluate $\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; {\mathrm{d}x}$

The key to evaluating this integral is to utilize Euler's reflection formula, which the proof can be looked up elsewhere, by substituting $u=1-x$ so that the Gamma function "disappears": $$I=\int_0^1 \ln{\left(\Gamma(1-u)\right)}\cos^2{(\pi u)} \; \mathrm{d}u$$ Now, add the original integral: \begin{align*} 2I&=\int_0^1 \ln{\left(\Gamma(x)\Gamma(1-x)\right)}\cos^2{(\pi x)} \; \mathrm{d}x \\ I&=\frac{1}{2} \int_0^1 \ln{\left(\frac{\pi}{\sin{(\pi x)}}\right)}\cos^2{(\pi x)} \; \mathrm{d}x \\ I&\overset{\pi x \to x}=\frac{1}{2 \pi} \int_0^{\pi} \ln{\left(\frac{\pi}{\sin{(x)}}\right)}\cos^2{(x)} \; \mathrm{d}x \\ &=\frac{\ln{\pi}}{2 \pi} \int_0^{\pi} \cos^2{(x)} \; \mathrm{d}x-\frac{1}{2 \pi} \int_0^{\pi} \cos^2{(x)} \ln{\left({\sin{(x)}}\right)} \; \mathrm{d}x \\ &= \frac{\ln{\pi}}{4}- \frac{1}{4 \pi}\underbrace{ \int_0^{\pi} \ln{(\sin{x})} \; \mathrm{d}x}_{I_1} - \frac{1}{4 \pi}\underbrace{ \int_0^{\pi} \cos{(2x)} \ln{(\sin{x})} \; \mathrm{d}x}_{I_2}\\ \end{align*}

Now, to calculate $I_1$, use symmetry and let $u=\frac{\pi}{2}-x$, then add the two integrals: \begin{align*} I_1&=\int_0^{\frac{\pi}{2}} \ln{(\sin{u})} +\ln{(\cos{u})}\; \mathrm{d}u \\ I_1&=\int_0^{\frac{\pi}{2}} \ln{(\sin{(2u)})}-\ln{2} \; \mathrm{d}u\\ I_1&=\frac{I_1}{2}-\frac{\pi\ln{2}}{2}\\ I_1 &= -\pi\ln{2}\\ \end{align*}

Now, to calculate $I_2$ \begin{align*} I_2&=2\int_0^{\frac{\pi}{2}} \cos{(2x)} \ln{(\sin{x})} \; \mathrm{d}x\\ &\overset{\sin{x} \to x}=2\int_0^1\frac{\left(1-2x^2\right)\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x \\ &=2\int_0^1\frac{\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x - 2\int_0^1 \frac{2x^2\ln{x}}{\sqrt{1-x^2}} \; \mathrm{d}x \\ &=-2\int_0^1 \frac{\arcsin{x}}{x} \mathrm{d}x+ 2\int_0^1 \frac{\arcsin{x}-2x\sqrt{1-x^2}}{x} \; \mathrm{d}x \\ &=-2\int_0^1 \frac{\arcsin{x}}{x} \; \mathrm{d}x+2\int_0^1\frac{\arcsin{x}}{x} \; \mathrm{d}x-2\int_0^1 \sqrt{1-x^2} \; \mathrm{d}x \\ &=-\frac{\pi}{2}\\ \end{align*}

Therefore, \begin{align*} \int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; \mathrm{d}x&=\frac{\ln{\pi}}{4}-\frac{1}{4\pi} \left(-\pi \ln{2}-\frac{\pi}{2}\right) \\ &= \boxed{\frac{\ln{(2\pi)}}{4}+\frac{1}{8}}\\ \end{align*}

Tags:

Integration

Calculus

Definite Integrals

Improper Integrals

Contour Integration

Related

How to show $\int_0^\infty\frac1{(1+x^2)(1+x^p)}$ doesn't depend on $p$? Olympiad question: In the regular pentagon $ABCDE$, the perpendicular at $C$ to $CD$ meets $AB$ at $F$. Prove that $AE + AF = BE$. How to evaluate $\int_{0}^{1} \! \frac{-\ln(1-t)}{t} \, \mathrm{d}t$ without using taylor series expansion? Generalizing Catalan numbers: number of ways where we cross the diagonal $k$ times. Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Probability of different winners in a two candidate election (range voting vs majority) If $M$ is a compact Riemannian manifold and $g$ and $\tilde{g}$ are metrics on $M$, then $\frac{1}{C} g \leq \tilde{g} \leq C g$ for $C > 1$ Gradient of $\mbox{dist}\left(x, D \right)^2:= \left\| x - P_{D}(x)\right\|_2^2$, where $P_{D}(x)$ is a projection operator Question related to continuity in Topology ( A proof i am unable to understand) For $W=\cup_{U\in\mathcal U} U$ show that there exists $U_1,\dots,U_n: \ \sum_{i=1}^n \lambda(U_i) > \frac{1 - \epsilon}{3^d}\lambda(W)$ $T:\Bbb{R}^2\rightarrow\Bbb{R}^2$ has 2 distinct eigenvalues. Showing that $v$ or $T(v)− \lambda_1v$ are eigenvectors of $T$ Coin Flip Problem

Recent Posts