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benorin

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10.Evaluate ##\int_0^{\infty}\frac{\log(x)}{x^2-1}dx.## (IR)

Owing to the infinite discontinuities of the integrand at ##x=0## and ##x=1##, I will split the integral up, $$I_1:=\int_0^{1}\frac{\log(x)}{x^2-1}dx\text{ and }I_2:=\int_1^{\infty}\frac{\log(x)}{x^2-1}dx$$

In ##I_1## substitute ##x=e^{-t}\implies dx=-e^{-t}dt\wedge 0\leq t \leq \infty## and the the integral becomes

$$\begin{gathered} I_1:=\int_0^{\infty}\frac{te^{-t}}{1-e^{-2t}}\, dt =\int_0^{\infty}t\sum_{k=0}^{\infty}e^{-(2k+1)t}\, dt \\ =\sum_{k=0}^{\infty}\int_0^{\infty}te^{-(2k+1)t}\, dt \\ \end{gathered} $$

let ##u=(2k+1)t## then we have

$$\begin{gathered} I_1=\sum_{k=0}^{\infty}\tfrac{1}{(2k+1)^2} \underbrace{\int_0^{\infty}te^{-u}\, dt}_{=\Gamma (2)=1} \\ =\sum_{k=0}^{\infty}\tfrac{1}{(2k+1)^2}=\lambda (2) \\ \end{gathered} $$

where ##\lambda (z)## is the Dirichlet Lambda Function.

The evaluation of the other integral, ##I_2##, goes the same as ##I_1## except that the first change of variable is ##x=e^t##, and the value is ##I_2 =\lambda (2)##. Hence the value of the given integral is ##2\lambda (2)##.