Trig Integral

Real Analysis
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Tolaso J Kos
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Trig Integral

#1

Post by Tolaso J Kos »

Show that: \( \displaystyle \int_{0}^{\pi/4}\ln \left ( \sin x \right )\ln \left ( \cos x \right )\left ( \frac{\ln \left ( \sin x \right )}{\cot x}+\frac{\ln \left ( \cos x \right )}{\tan x} \right )\, dx=-\frac{\pi^4}{2880} \).
Imagination is much more important than knowledge.
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Tolaso J Kos
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Re: Trig Integral

#2

Post by Tolaso J Kos »

Solution by Omran Kouba.

Let \( \displaystyle J=\int_{0}^{\pi/2}\ln^2(\sin x) \ln (\cos x) \tan x\,dx \).
Then: $$\begin{aligned}
J &=\int_{0}^{\pi/4}\ln^2\left ( \sin x \right ) \ln \left ( \cos x \right )\tan x\,dx+\int_{\pi/4}^{\pi/2}\ln^2\left ( \sin x \right )\ln \left ( \cos x \right )\tan x\,dx\\
&= \int_{0}^{\pi/4}\ln^2\left ( \sin x \right ) \ln \left ( \cos x \right )\tan x\,dx+\int_{0}^{\pi/4}\ln^2 \left ( \cos x \right )\ln \left ( \sin x \right )\cot x\,dx\\
&= I\\
\end{aligned}$$ To calculate \( J \) we apply the sub \( t=\sin^2 x \) thus \( \displaystyle J=\frac{1}{16}\int_{0}^{1}\frac{\ln \left ( 1-u \right )\ln^2 u}{1-u}\,du \).
Since \( \displaystyle \frac{\ln \left ( 1-u \right )}{1-u}=-\left ( \sum_{n=0}^{\infty}u^n \right )\left ( \sum_{n=1}^{\infty}\frac{u^n}{n} \right )=-\sum_{n=1}^{\infty}\mathcal{H}_nu^n \) the previous equation is re-written as: $$J=-\frac{1}{16}\sum_{n=1}^{\infty}\mathcal{H}_n\int_{0}^{1}u^n\ln^2 u\,du=-\frac{1}{8}\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{\left ( n+1 \right )^3}$$
It also holds: \( \displaystyle \mathcal{H}_n=\mathcal{H}_{n+1}-\frac{1}{n+1} \).
So our last equation is re-written as:
$$J=\frac{1}{8}\zeta(4)-\frac{1}{8}\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^3}\quad (*)$$
All sums are known:

Thus: $$\boxed{\displaystyle J=\frac{1}{8}\zeta(4)-\frac{1}{16}\zeta^2(2)=-\frac{\pi^4}{2880}}$$
Imagination is much more important than knowledge.
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Tolaso J Kos
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Re: Trig Integral

#3

Post by Tolaso J Kos »

\( (*) \)
Proof:
$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^3} &=\sum_{n=1}^{\infty}\frac{1}{n^3}\int_{0}^{1}\frac{1-t^n}{1-t}\,dt \\
&= \int_{0}^{1}\frac{1}{1-t} \sum_{n=1}^{\infty}\frac{1-t^n}{n^3}\,dt\\
&= \int_{0}^{1}\frac{\zeta(3)-{\rm Li_3}(t)}{1-t}\,dt
\end{aligned}$$

Integrating by parts we get:
$$\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^3}=-\left ( \zeta(3)-{\rm Li_3}(t) \right )\ln(1-t)\bigg|_0^1-\int_{0}^{1}\frac{\ln (1-t){\rm Li_2}(t)}{t}\,dt$$
The first terms goes to zero. Thus:
$$\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^3}=-\int_{0}^{1}\frac{\ln(1-t){\rm Li_2}(t)}{t}\,dt=\left [ \frac{{\rm Li^2_2(t)}}{2} \right ]_0^1=\frac{\zeta^2(2)}{2}=\frac{\pi^4}{72}$$

This is a well known formula and there are plenty of proofs around.

Euler has also proved in \( 1775 \) that:
$$2\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^q}=\left ( q+2 \right )\zeta(q+1)-\sum_{m=1}^{q-2}\zeta(m+1)\zeta(q-m)$$
Imagination is much more important than knowledge.
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