calculus - Closed form of $int_{0}^{infty} frac{tanh(x),tanh(2x)}{x^2};dx$

I have homework to evaluate this integral

$$I=\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx$$

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Here is what I have done so far. I tried integration by parts using $u=\tanh(x)\,\tanh(2x)$ and $dv=\frac{dx}{x^2}$, I got
$$\begin{align}\int_{0}^{\infty} \frac{\tanh(x)\,\tanh(2x)}{x^2}\;dx&=-\left.\frac{\tanh(x)\,\tanh(2x)}{x}\right|_{0}^{\infty}+2\int_{0}^{\infty}\frac{\tanh(2x)\,\text{sech}(2x)}{x}\;dx\\&=2\int_{0}^{\infty}\frac{\tanh(2x)\,\text{sech}(2x)}{x}\;dx\end{align}$$
At this part I'm stuck. I'm thinking of using Frullani's integral but I'm having trouble to find a relation as such $\tanh(2x)\,\text{sech}(2x)=f(ax)-f(bx)$.

I also tried using differentiation under integral sign by considering
$$I(a,b)=\int_{0}^{\infty} \frac{\tanh(ax)\,\tanh(bx)}{x^2}\;dx$$

then
$$\frac{dI}{da}=\int_{0}^{\infty} \frac{\text{sech}^2(ax)\,\tanh(bx)}{x}\;dx=\int_{0}^{\infty} \frac{\tanh(bx)-\tanh^2(ax)\tanh(bx)}{x}\;dx$$
Again I tried to use Frullani's integral but I'm having trouble to find the sufficient $f(x)$. Integrating again with respect to $b$, I got
$$\frac{d^2I}{da\;db}=\int_{0}^{\infty} \text{sech}^2(ax)\text{sech}^2(bx)\;dx$$
It's obviously a dead end to me. At this rate my friends and I contacted my professor to confirm whether the integral can be evaluated in terms of elementary functions or not because W|A cannot find it (I know that W|A cannot do everything). He only said, "Sure! The answer is only 3 characters" and then he left us. Assuming he is right, so $I$ must have a nice closed form, but I'm unable to find it.

Would you help me? Any help would be appreciated. Thanks in advance.

Answer

Your professor is right. Note that
$$ \tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}, \tanh(2x)=\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}=\frac{(e^x-e^{-x})(e^x+e^{-x})}{e^{2x}+e^{-2x}}$$

and hence
\begin{eqnarray*}
\int_0^\infty\frac{\tanh(x)\tanh(2x)}{x^2}dx&=&\int_0^\infty\frac{(e^{x}-e^{-x})^2}{x^2(e^{2x}+e^{-2x})}dx\\
&=&\int_0^\infty\frac{e^{2x}-2+e^{-2x}}{x^2(e^{2x}+e^{-2x})}dx.
\end{eqnarray*}
Now define
$$ I(a)=\int_0^\infty\frac{e^{ax}-2+e^{-ax}}{x^2(e^{2x}+e^{-2x})}dx$$
to get
\begin{eqnarray}
I''(a)&=&\int_0^\infty\frac{e^{(-a-2)x}+e^{(-a+2)x}}{1+e^{-4x}}dx,\\

&=&\int_0^\infty\sum_{n=0}^\infty(-1)^n(e^{(-a-2)x}+e^{(-a+2)x})e^{-4nx}dx\\
&=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\
&=&\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right).
\end{eqnarray}
So
\begin{eqnarray}
I'(a)&=&\int_0^a\frac{\pi}{4}\sec\left(\frac{\pi t}{4}\right)dt\\
&=&\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)-\ln\cos\left(\frac{\pi a}{4}\right)
\end{eqnarray}
and hence

\begin{eqnarray}
I(2)&=&\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da-\int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da.
\end{eqnarray}
Note
$$ \int_0^2\ln\cos\left(\frac{\pi a}{4}\right)da=-2\ln2 $$
from Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $
and it should not be hard to get
$$ \int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$$
and thus
$$ I(2)=\frac{8G}{\pi}. $$

$\bf{Update}$ 1: Let us first work on
$ \sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)=\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$.
In fact
\begin{eqnarray*}
&&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{4n-a+2}+\frac{1}{4n+a+2}\right)\\
&=&\sum_{n=0}^\infty\left(\frac{1}{8n-a+2}-\frac{1}{8n-a+6}\right)+\sum_{n=0}^\infty\left(\frac{1}{8n+a+2}-\frac{1}{8n+a+6}\right)\\
&=&\sum_{n=0}^\infty\frac{4}{(8n-a+2)(8n-a+6)}+\sum_{n=0}^\infty\frac{4}{(8n+a+2)(8n+a+6)}\\
&=&\sum_{n=-\infty}^\infty\frac{4}{(8n-a+2)(8n-a+6)}=\sum_{n=-\infty}^\infty\frac{4}{(8n-a+4)^2-2^2}\\
&=&\frac{1}{16}\sum_{n=-\infty}^\infty\frac{1}{(n+\frac{4-a}{8})^2-(\frac{1}{4})^2}.

\end{eqnarray*}
Now using a result from Closed form for $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}$ for $a=\frac{i}{4}$ and $z=\frac{4-a}{8}$ and after some basic calculation we can get this result. Also see An alternative proof for sum of alternating series evaluates to $\frac{\pi}{4}\sec\left(\frac{a\pi}{4}\right)$ for a short proof.

$\bf{Update}$ 2: We work on $\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{8G}{\pi}-2\ln 2$. In fact
\begin{eqnarray}
\int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln\left(1+\sin(a)\right)da=\frac{4}{\pi}\int_0^{\pi/2}\ln(1+\cos(a))da.
\end{eqnarray}
Using $2\cos^2\frac{a}{2}=1+\cos a$ and a result $G=\int_0^{\pi/4}\ln(\cos(t))dt$ from http://en.wikipedia.org/wiki/Catalan%27s_constant, it is easy to obtain
$$ \int_0^2\ln\left(1+\sin\left(\frac{\pi a}{4}\right)\right)da=
\frac{8G}{\pi}-2\ln 2. $$