Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$
Proof. The function
$$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$
is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that
$$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$
If we make change of variables in the latter integral $t\to 1-t$ we obtain
$$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$
On the other hand, using integration by parts, we obtain
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$
Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence
$$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$
Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain
$$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$
On the other hand $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t+\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t=\int_{0}^1\frac{\log(1-t)}{t}\text{d} t=-\frac{\pi^2}{6}$$
Substracting the last two equalities we obtain $$\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d}
t=-\frac{\pi^2}{12}-\frac{1}{2}\log\left(\frac{1}{2}\right)^2$$
and thus $\displaystyle S\left(\frac{1}{2}\right)=\frac{\pi^2}{12}+\frac{1}{2}\log\left(\frac{1}{2}\right)^2$.
Some math jokes (which turn out to be correct)
Prove that $$\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{\pi}+1)}\text{d}x=\int_{0}^{\infty}\frac{1}{(x^2+1)(x^{e}+1)}\text{d}x$$
Prove that $$\int_{0}^{\pi/2}\frac{1}{1+\tan^{\pi}(x)}\text{d}x=\int_{0}^{\pi/2}\frac{1}{1+\tan^{e}(x)}\text{d}x$$
$\textbf{Proof.}$ Both of the integrals are being solved, when the peculiar powers are being replaced by a parameter "a" and then differentiation with respect to "a" is performed. Actually the integrals are related vie the change of variables $x\to\arctan(x)$.
For the first integral, denote
$$f(x,a)= \frac{1}{(x^2+1)(x^{a}+1)}.$$
Then $$\frac{\partial }{\partial a}f(x,a)=\frac{x^a \log (x)}{\left(x^a+1\right) \left(x^a+1\right)^2}$$
Consider the integral
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x.$$ Making the change $x\to 1/x$ this transforms to the integral
$$\int_{\infty}^{1}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{-1}{x^2}\text{d}x=\int_{1}^{\infty}\frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}\text{d}x$$
Now it remains to observe that
$$ \frac{\partial }{\partial a}f\left(\frac{1}{x},a\right)\frac{1}{x^2}=-\frac{\partial }{\partial a}f(x,a)$$
so that
$$\int_{0}^{1}\frac{\partial }{\partial a}f(x,a)\text{d}x=-\int_{1}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x$$
and hence
$$\int_{0}^{\infty}\frac{\partial }{\partial a}f(x,a)\text{d}x=0.$$
Thus $$\int_{0}^{\infty}f(x,a)\text{d}x$$
is a constant independent of $a$, which proves the equality. Plugging $a=0$ we see moreover that the value of the integrals is $\displaystyle \frac{\pi}{4}$.
$\textbf{Addendum}$. Another solution proceeds as directly making the change of variable $x\to 1/x$. Thus
$$\int_0^\infty\frac{1}{(x^2+1)(x^{a}+1)}\text{d}x=\int_0^\infty\frac{x^a}{(x^2+1)(x^{a}+1)}\text{d}x.$$
Summing both integrals yields the solution.
The second problem could be done in a similar manner.
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