# Evaluating $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ using complex analysis

Again, improper integrals involving $\ln(1+x^2)$

I am trying to get a result for the integral $I_{\alpha}=\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}dx$ - asked above link- using some complex analysis, however, I couldn't find an appropriate solution. (Of course, for $\alpha \in (0,3)$ as stated above link (o.w. it is divergent))

Any ideas?

• Applying by parts with $u=ln(1+x^2)$ and $v'=x^{- \alpha}$ we have: $\int uv'= x^{-1-\alpha} ln(1+x^2) - \int 2dx/((1+x^2)x^{\alpha})$ then I guess branch cut? Nov 16, 2014 at 19:52

Let $\alpha = 1 + p$. Then we have $$\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx$$ where $0<p<2$. Let $u = \ln(1+x^2)$ and $dv = x^{-1-p}dx$. Then $du = \frac{2x}{1+x^2}$ and $v = \frac{-1}{px^p}$. $$uv\big|_0^{\infty}\to 0$$ for $p\in(0,2)$. $$\int_0^{\infty}\frac{\ln(1+x^2)}{x^{1+p}}dx=\frac{2}{p}\int_0^{\infty}\frac{x^{1-p}}{x^2+1}dx$$ Let $\beta -1 = 1-p$ so $p=2-\beta$. Let $x=t^2$ so $dx=2tdt$. $$\frac{4}{2-\beta}\int_0^{\infty}\frac{t^{2\beta - 1}}{t^4+1}dt$$ The contour we desire is

The integrals over the semicircles go to zero as $R\to\infty$ and $\delta\to 0$ so we have \begin{align} \frac{4}{2-\beta}\int_0^{\infty}\frac{t^{2\beta - 1}}{t^4+1}dt &= \frac{4}{2-\beta}\int_{-\infty}^{\infty}\frac{z^{2\beta - 1}}{z^4+1}dz\\ &= \frac{4}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}+(-z)^{2\beta - 1}}{z^4+1}dz \end{align} Now, $(-z)^{2\beta} = -e^{2\pi i\beta}z^{2\beta}$. $$\frac{4}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}+(-z)^{2\beta - 1}}{z^4+1}dz = \frac{4(1-e^{2\pi i\beta})}{2-\beta}\int_0^{\infty}\frac{z^{2\beta - 1}}{z^4+1}dz\tag{1}$$ Then $(1)$ is equal to $2\pi i$ times sum of the residue is the upper half plane. \begin{align} \frac{8\pi i}{(1-e^{2\pi i\beta})(2-\beta)}\Bigl[\lim_{z\to e^{\frac{i \pi }{4}}}\frac{(z-e^{\frac{i \pi }{4}}) z^{2\beta -1}}{z^4+1}+\lim_{z\to e^{\frac{3i\pi}{4}}}\frac{(z-e^{\frac{3i\pi}{4}})z^{2\beta-1}}{z^4+1}\Bigr] &= \frac{\pi\csc\bigl(\frac{\pi\beta}{2}\bigr)}{2-\beta}\\ &= \frac{\pi\csc\bigl(\frac{\pi p}{2}\bigr)}{p}\tag{2}\\ &= \frac{\pi\sec\bigl(\frac{\pi\alpha}{2}\bigr)}{1-\alpha}\tag{3} \end{align} Equations $(2)$ and $(3)$ occur by back substituting for $\beta$ and $p$.

Differentiation under the integral sign is another possibility. Let: $$I(\alpha,\beta) = \int_{0}^{+\infty}\frac{\log(1+\beta x^2)}{x^\alpha}\,dx. \tag{1}$$ Then assuming $1<\alpha<3$ we have: $$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+\beta x^2}\,dx = \beta^{\frac{\alpha-3}{2}}\int_{0}^{+\infty}\frac{x^{2-\alpha}}{1+x^2}\,dx\tag{2}$$ and by replacing $x$ with $\tan\theta$ we get, through the Euler beta function and the reflection formula for the $\Gamma$ function: $$\frac{\partial}{\partial\beta} I(\alpha,\beta) = \beta^{\frac{\alpha-3}{2}}\int_{0}^{\pi/2}\left(\tan\theta\right)^{2-\alpha}\,d\theta = -\beta^{\frac{\alpha-3}{2}}\frac{\pi}{2\cos\left(\frac{\pi \alpha}{2}\right)}\tag{3}$$ and by integrating $(3)$ with respect to $\beta$ we have: $$\int_{0}^{+\infty}\frac{\log(1+x^2)}{x^\alpha}\,dx = \frac{\pi}{1-\alpha}\,\sec\left(\frac{\pi\alpha}{2}\right).\tag{4}$$

• @dustin: I did not feel insulted in any way, I was just working on a completely different problem. And upvoting is not compulsive for sure :) Mar 29, 2015 at 21:38