\(\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2(x-\frac{\pi}{4})}\, {\rm d}x\)
- Grigorios Kostakos
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\(\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2(x-\frac{\pi}{4})}\, {\rm d}x\)
Evaluate \[\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2\big(x-\frac{\pi}{4}\big)}\, {\rm d}x\,.\]
Grigorios Kostakos
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Re: \(\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2(x-\frac{\pi}{4})}\, {\rm d}x\)
Hmm this seems to have been forgotten. Anyway, here is an answer.Grigorios Kostakos wrote:Evaluate \[\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2\big(x-\frac{\pi}{4}\big)}\, {\rm d}x\,.\]
\begin{align*}
\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\sin^2 \left ( x- \frac{\pi}{4} \right )} \, {\rm d}x &= \int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right )^2} \, {\rm d}x\\
&= 2\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \sin x - \cos x \right )^2} \, {\rm d}x\\
&= 2 \int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \frac{\sin x}{ \cos x} -1 \right )^2 \cos^2 x} \, {\rm d}x\\
&= 2\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \tan x - 1 \right )^2} \sec^2 x \, {\rm d}x\\
&\!\!\!\!\! \overset{u=\tan x}{=\! =\! =\! =\!} \; 2 \int_{0}^{\infty} \frac{\log^2 u}{\left ( u-1 \right )^2} \, {\rm d}u \\
&= 2 \left [ \int_{0}^{1} \frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d} u + \int_{1}^{\infty} \frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d}u \right ] \\
&=4 \int_{0}^{1}\frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d}u \\
&= 4 \int_{0}^{1} \log^2 u \sum_{n=1}^{\infty} n u^{n-1} \, {\rm d}u \\
&= 4 \sum_{n=1}^{\infty} n \int_{0}^{1} u^{n-1 }\log^2 u \, {\rm d}u \\
&= 4 \cdot 2 \sum_{n=1}^{\infty} \frac{1}{n^2} \\
&= 8 \zeta(2) \\
&=\frac{4\pi^2}{3}
\end{align*}
Imagination is much more important than knowledge.
Re: \(\int_{0}^{\frac{\pi}{2}} \frac{\log^2(\tan{x})}{\sin^2(x-\frac{\pi}{4})}\, {\rm d}x\)
Hmm.. we can make a somewhat generalization for this integral. Let us consider $\mathbb{N} \ni n \geq 2$ as well as the family of integrals:
$$\mathcal{J}(n)=\int_0^{\pi/2} \frac{\log^n \tan x }{\sin^2 \left( x - \frac{\pi}{4} \right)} \, {\rm d}x$$
Then following the exact steps of Tolaso we get the following:
$$\mathcal{J}(n)=\left\{\begin{matrix}
0& , & \text{if } \; n \; \text{is odd}\\
4n! \zeta(n)& , & \text{if } \; n \; \text{is even}
\end{matrix}\right.$$
Unfortunately, we cannot play with the square of the denominator due to convergence issues.
Something that may explain why $0$ pops up if $n$ is odd may be found here .
$$\mathcal{J}(n)=\int_0^{\pi/2} \frac{\log^n \tan x }{\sin^2 \left( x - \frac{\pi}{4} \right)} \, {\rm d}x$$
Then following the exact steps of Tolaso we get the following:
$$\mathcal{J}(n)=\left\{\begin{matrix}
0& , & \text{if } \; n \; \text{is odd}\\
4n! \zeta(n)& , & \text{if } \; n \; \text{is even}
\end{matrix}\right.$$
Unfortunately, we cannot play with the square of the denominator due to convergence issues.
Something that may explain why $0$ pops up if $n$ is odd may be found here .
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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