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How to evaluate the following integral/ $$\int_{0}^{\infty} \, \frac{x^\alpha \log^n{(x^\beta)}}{k^m+x^m}\, dx $$, $\alpha, \beta, m, n$ are real and $k$ is a positive integer.

For the case where $\beta=1$ and $m=2$, we can consider the following integral: $$ \begin{aligned} I(\alpha, k) & =\int_0^{\infty} \frac{x^\alpha}{k^2+x^2} d x \\ & =\frac{k^{\alpha-1}}{2} \int_0^{\infty} \frac{t^{(\alpha-1) / 2}}{1+t} d t \\ & =\frac{k^{\alpha-1}}{2} \mathrm{~B}\left(\frac{1+\alpha}{2}, \frac{1-\alpha}{2}\right) \\ & =\frac{k^{\alpha-1}}{2} \frac{\pi}{\cos \left(\frac{\pi}{2} a\right)} \end{aligned} $$

Differentiating $I(\alpha, k)$ with respect to $\alpha$ yields: $$ \begin{align*} \int_{0}^{\infty} \, \frac{x^\alpha \log^n{x}}{k^2+x^2}\, dx &= \frac{\partial^{n} }{\partial a^n} \left(\frac{k^{\alpha-1}}{2}\, \frac{\pi}{\cos{\displaystyle \left(\frac{\pi}{2}\alpha\right)}}\right) \end{align*} $$(based on this answer).

Are there any other methods to evaluate the general integral? Thank you!

J.G.
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Anomaly
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1 Answers1

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Starting from your integral $$J= \int\limits_0^\infty \frac{x^\alpha \left(\ln x^\beta\right)^n}{k^m +x^m}\, dx, \quad 0\lt 1-\alpha \lt m, \quad k\gt 0, \quad n\in \mathbb{N}, $$ the substitution $x^m =k^m u$ leads to $$J=\frac{k^{1+\alpha-m} \beta^n}{m^{n+1}}\sum\limits_{\ell=0}^n \left(\begin{array} &n \\[-4pt] \ell \end{array} \right) \left( \ln k^m\right)^{n-\ell} \int\limits_0^\infty \frac{\,u^{-1+(1+\alpha)/m}\, (\ln u)^\ell}{1+u}\, du, $$ being a linear combination of the simpler integrals$$I_\ell(\gamma):= \int\limits_0^\infty \!\frac{\, u^{-\,\gamma}\, \left( \ln u \right)^\ell}{1+u} \, du, \quad 0\lt \gamma \lt 1, \; \ell \in \mathbb{N}.$$ The "master integral" $$ I_0(\gamma)=\int\limits_0^\infty \! \frac{u^{-\, \gamma}}{1+u}\,du =\frac{\pi}{\sin \pi \gamma}$$ is easily computed by complex integration employing a keyhole contour, with all other integrals given by $$I_\ell(\gamma) = (-1)^\ell \, \frac{d^\ell I_0(\gamma)}{d \gamma^\ell} .$$

Hyperon
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