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Once I encountered the integral

$$\int_0^{\frac{\pi}{2}} \sin (2 x) \ln ^2\left(1+2 \tan ^2 x\right) d x ,$$

my first substitution is $c=\cos^2 x$

$\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \sin (2 x) \ln ^2\left(1+2 \tan ^2 x\right) d x = & -\int_0^{\frac{\pi}{2}} \ln ^2\left(\frac{\cos ^2 x+2 \sin ^2 x}{\cos ^2 x}\right) d\left(\cos ^2 x\right) \\ = & \int_0^1 \ln ^2\left(\frac{2-c}{c}\right) d c\end{aligned}\tag*{} $

Then I put $y=\frac{2-c}{c}$ and get

$\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \sin (2 x) \ln ^2\left(1+2 \tan ^2 x\right) d x & =2 \int_1^{\infty} \frac{\ln ^2 y}{(1+y)^2} d y \\ & =-2 \int_1^{\infty} \ln ^2 y d\left(\frac{1}{1+y}\right) \\ & =4 \int_1^{\infty} \frac{\ln y}{y(1+y)} d y \quad \textrm{ (Via integration by parts)} \\ & =4 \int_0^1 \frac{-\ln y}{\frac{1}{y}\left(1+\frac{1}{y}\right)} \frac{d y}{y^2}, \text { where } y \mapsto \frac{1}{y} \\ & =-4 \int_0^1 \frac{\ln y}{y+1} d y \\ & =-4\left(-\frac{\pi^2}{12}\right) \quad \textrm{ (See footnote for details)} \\ & =\frac{\pi^2}{3}\end{aligned}\tag*{} $

Then I investigate the integral in general

$$I_n=\int_0^{\frac{\pi}{2}} \sin (2 x) \ln ^n\left(1+2 \tan ^2 x\right) d x ,$$

Using the same substitution $y=\frac{2-\cos^2x}{\cos^2 x}$, we get

$$ \begin{aligned} I_n & =2 n \int_1^{\infty} \frac{\ln ^{n-1} y}{y(1+y)} d y \\ & =2 n \int_0^1 \frac{\ln ^{n-1}\left(\frac{1}{y}\right)}{\frac{1}{y}\left(1+\frac{1}{y}\right)} \frac{d y}{d y^2} \\ & =2 n(-1)^{n-1} \int_0^1 \frac{\ln ^{n-1} y}{1+y} d y\\&=2^{1-n} n\left(2^n-2\right) \zeta(n) \Gamma(n) \end{aligned} $$

where the last answer using the result in my post.

My question:

Is there any other method to evaluate $\int_0^{\frac{\pi}{2}} \sin (2 x) \ln ^n\left(1+2 \tan ^2 x\right) d x ?$

Footnote:

$\displaystyle \begin{aligned} \int_0^1 \frac{\ln y}{y+1} d y & =\sum_{n=0}^{\infty}(-1)^n \int_0^1 y^n \ln y d y=\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \\ & =-\sum_{n=1}^{\infty} \frac{1}{n^2}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n)^2}=-\frac{1}{2} \sum_{n=1}^n \frac{1}{n^2}\\&=-\frac{\pi^2}{12}\end{aligned}\tag*{} $

Lai
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1 Answers1

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Recall Dirichlet Eta Function $\eta(s)\,$: $$ \begin{align} &\Gamma(s)\,\eta(s) =\left(1-2^{1-s}\right)\,\Gamma(s)\,\zeta(s) =\int_{0}^{\infty}\frac{y^{s-1}}{e^y+1}\,dy \\[2mm] &\qquad\color{purple}{{\small\text{IBP}}\begin{cases}v=1/\left(e^y+1\right)&\implies dv=-e^y/\left(e^y+1\right)^2\,dy \\ du=y^{s-1}\,dy &\implies u=y^s/s\end{cases}} \\[2mm] &\Gamma(s)\,\eta(s) =\left[\frac{y^s/s}{e^y+1}\right]^{\infty}_{0} +\int_{0}^{\infty}\frac{e^y\,y^s/s}{\left(e^y+1\right)^2}\,dy =\frac{1}{s}\int_{0}^{\infty}\frac{y^s\,e^y}{\left(e^y+1\right)^2}\,dy \end{align} $$ Hence, $$ \begin{align} &I_s =\int_{0}^{\pi/2}\ln^s\left(1+2\tan^2x\right)\,\sin(2x)\,dx \\[2mm] &\color{blue}{\ln\left(1+2\tan^2x\right)=y} \implies \color{blue}{\tan^2x=\frac{1}{2}\left(e^y-1\right)}\,,\,{\small{\int_0^{\pi/2}\,dx \mapsto \int_{0}^{\infty}\,dy}} \\[2mm] &2\frac{\sin x}{\cos^3x}\,dx=\frac{1}{2}e^y\,dy \implies 2\sin x\cos x\,dx=\frac{1}{2}\left(\cos^2x\right)^2e^y\,dy \\[2mm] &\color{blue}{\sin(2x)\,dx} =\frac{1}{2}\left(\frac{1}{1+\tan^2x}\right)^2e^y\,dy =\color{blue}{\frac{2\,e^y\,dy}{\left(e^y+1\right)^2}} \\[4mm] &\color{red}{I_s} =2\int_{0}^{\infty}\frac{y^s\,e^y}{\left(e^y+1\right)^2}\,dy =\color{red}{2s\,\Gamma(s)\,\eta(s)} \\[2mm] &\quad=2s\left(1-2^{1-s}\right)\,\Gamma(s)\,\zeta(s)\,:\,\forall s\in\mathbb{C}\land\,\Re(s)\gt0 \end{align} $$

Hazem Orabi
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