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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 ot

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(\