TITLE: How evaluate $ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2} \;dx$
QUESTION [9 upvotes]: How evaluate (without using Complex analysis)
$$ \int_0^\infty \frac{\ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|}{1+x^2}\; dx\quad (a\gt0)$$
My Attempt:
I used the expansion of the following functions: $$ \ln\left|2\sin(x)\right| \text{ and }\ln\left|2\cos(x)\right| $$
To get the following expansion: $$ \ln\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=2\sum_{n=1}^\infty (-1)^n \frac{\sin[(2n-1)ax]}{2n-1} $$
Then I expressed the square of the logarithmic function as follows:$$ \ln^2\;\left|\;\tan\left(\frac{ax}{2}-\frac{\pi}{4}\right)\;\right|=4\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{m+n} \frac{\sin[(2m-1)ax]\sin[(2n-1)ax]}{(2m-1)(2n-1)} $$
And used the formula of the product of two sines, then integrated the following well known integral under the summation sign:
$$ \int_0^\infty \frac{\cos(bx)}{1+x^2} \;dx$$
And expressed the final result in terms of $\; \arctan\;\left(\;e^{-a}\;\right)\; $
but this was not the right answer according to the one evaluated by Complex analysis, which is
$$ \frac{{\pi}^3}{8}-2\pi\; \arctan^2\;\left(\;e^{-a}\;\right) $$
Any hint for another method or idea?

REPLY [2 votes]: part(1)
$$first\ part\\
I=\int_{0}^{\infty }\frac{ln^2(tan(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2}-\frac{\pi }{4}))}{x^2+1}dx\\
\\
we\ have\ ln^2(cot(\frac{ax}{2}-\frac{\pi }{4})=-2\sum_{n,m=1}^{\infty }(-1)^{m+n}\frac{sin(2m-1)ax.sin(2n-1)ax}{(2m-1)(2n-1))}\\
\\
 =2\sum_{n,m=1}^{\infty }\frac{(-1)^{m+n}[cos(2m-2n)ax-cos(2m+2nax))]}{(2m-1)(2n-1)}\\
\\
\therefore I=\frac{\pi }{2}\sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}[e^{-2a(2m-2n)}-e^{-2a(2m+2n-2)}]\\
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note\ \sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}e^{-2a(2n+2m-2)}\\
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=\sum_{m=1}^{\infty }\frac{(-1)^{m-1}}{(2m-1)}e^{-2a(2m-1)}.\sum_{n=1}^{\infty }\frac{(-1)^{n-1}}{(2n-1)}e^{-2a(2n-1)}=arctan^2(e^{-2a})$$
second  part 
$$second\ part \\
\because \sum_{m,n=1}^{\infty }\frac{(-1)^{m+n}}{(2m-1)(2n-1)}e^{-2a(2m-2n)}=\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}+\sum_{n,m=1}^{\infty }\frac{(-1)^{m-n}}{(2m-1)(2n-1)}e^{-2a(2m-2n)}\\
\\
=\frac{\pi ^{3}}{8}+\sum_{m,n=1}^{\infty }\frac{(-1)^{m-n}}{(m-n)(2n-1)}e^{-2a(2m-2n)}=\frac{\pi ^{3}}{8}+\sum_{n>=1}^{\infty }\frac{1}{(2n-1)}.\sum_{n,m>=1}^{\infty }\frac{(-1)^{m-n}}{(m-n)}e^{-2a|2m-2a}\\
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=\frac{\pi ^{3}}{8}+\sum_{n=1}^{\infty }\frac{1}{2n-1}.\sum_{k=n}^{\infty }e^{-4a|k|}\Rightarrow k=m-n\\
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=\frac{\pi ^{3}}{8}+\sum_{k=n}^{\infty }\sum_{n=1}^{\infty }\frac{1}{2n-1}\frac{(-1)^{k}}{k}e^{-4ak}=\frac{\pi ^{3}}{8}+\sum_{k=1}^{\infty }[\sum_{n=1}^{k}\frac{1}{2n-1}]\frac{(-1)^{k}}{k}e^{-4ak}$$
part(3)
$$third \ part\\
but\ 2\sum_{n=1}^{k}\frac{1}{2n-1}=\sum_{i=1}^{k}\frac{1}{2i-1}+\sum_{g=1}^{k}\frac{1}{2g-1}\\
\\
=\sum _{i+g=k+1}(\frac{1}{2i-1}+\frac{1}{2g-1})=\sum _{i+g=k+1}\frac{2(i+g-1)}{(2i-1)(2g-1))}\\
\therefore \sum_{n=1}^{k}\frac{1}{2n-1}=\sum _{i+g=k+1}\frac{k}{(2i-1)(2g-1))}\\
\therefore \sum _{k>=1}[\sum _{i+g=k+1}\frac{k}{(2g-1)(2i-1)}]e^{-2a(2m-2n)}=-\sum _{i,g>=1}\frac{(-1)^{i-g}}{(2i-1)(2g-1)}e^{-2a(2i+2g-2)}\\
\\
=-arctan^2(e^{-2a})\ \ \ \ \ \ so \ we \ have\  I=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2}))-\frac{\pi }{4})}{x^2+1}dx\\
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=\frac{\pi ^{3}}{16}-\pi arctan^2(e^{-2a})\\
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\therefore I=\int_{0}^{\infty }\frac{ln^2(cot(\frac{ax}{2})-\frac{\pi }{4})}{x^2+4}dx =\frac{\pi ^{3}}{8}-2\pi arctan^2 (e^{-2a})\\
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\ \ \ \ \ \ \ \ \ \ ahmed\ hegazi$$