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A logarithmic integral

Prove that

\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \log x}{\left ( 1+x \right )^2} \, {\rm d}x = \pi

Solution

We begin by making the substitution u=\sqrt{x} thus:

\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \log x}{(1+x)^2} \, {\rm d}x \overset{u=\sqrt{x}}{=\! =\! =\! =\!} \int_{0}^{\infty} \frac{4u^2 \log u}{\left ( 1+u^2 \right )^2} \, {\rm d}u

Now let us consider the complex function \displaystyle f(z)=\frac{z^2 \log^2 z}{(1+z^2)^2} where the principal arguement of z lies within the interval (-\pi, \pi] as well as the contour below

It is clear that f has two poles of order 2 at z=2i and z=-2i. The residue at z=i is equal to  \frac{\pi}{4} + \frac{i \pi^2}{16} whereas the residue at z=-i is equal to  \frac{\pi}{4} - \frac{i \pi^2}{16} . Thus

\displaystyle \oint \limits_{\gamma} f(z) \, {\rm d}z = \oint \limits_{\mathcal{C}_R} f(z) \, {\rm d}z + \oint \limits_{\mathcal{C}_\epsilon} f(z) \, {\rm d}z + \int_{-R}^{-\epsilon} f(z) \, {\rm d}z + \int_{-\epsilon}^{-R} f(z) \, {\rm d}z

Sending \epsilon \rightarrow 0 and R \rightarrow + \infty the contribution of both the large and the small circle is 0. Hence:

\begin{aligned} i \pi^2 &= \int_{-\infty}^{0} \frac{x^2 \left ( \log \left | x \right | + i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x + \int_{0}^{-\infty} \frac{x^2 \left ( \log \left | x \right | - i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &\!\!\!\!\! \overset{y=-x}{=\! =\! =\! =\! =\!} \int_{0}^{\infty} \frac{x^2 \left ( \log x + i \pi \right )^2}{\left ( 1+x^2 \right )^2} - \int_{0}^{\infty} \frac{x^2 \left ( \log x - i \pi \right )^2}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &=\bigintss_{0}^{\infty} \frac{x^2 \bigg( \left ( \log x + i \pi \right )^2 - \left ( \log x - i \pi \right )^2 \bigg)}{\left ( 1+x^2 \right )^2} \, {\rm d}x \\ &= 4 i \pi \int_{0}^{\infty} \frac{x^2 \log x}{\left ( 1+x^2 \right )^2} \, {\rm d}x \end{aligned}

Thus the conclusion follows.

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