Any general formulas for evaluating integrals of the form $ \int_{-\infty}^{\infty} \frac{1}{(x^2+k^2)^n} \,dx $

Question

I was working one some integrals in my spare time, and I was wondering, is there any general formula for evaluating these integrals for any $n>1/2$ and any real $k$.

$$ \int_{-\infty}^{\infty} \frac{1}{(x^2+k^2)^n} \,dx $$

I tried looking on the internet, but only found something for the special case of $k=1$, but I was wondering if it possible to generalize for any $k$. I don't mind formula, if it exists, containing some special functions, like Gamma, or others, but I was curious if there is any kind of a formula.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Mar 28 '25 at 22:11
Once you know the result for $k=1$ you know the result for all $k$ — Sine of the Time, Mar 28 '25 at 22:12

score 1 · Answer 1 · edited Mar 28 '25 at 23:29

$$ \begin{align} \int_{-\infty}^\infty \frac{\mathrm dx}{(x^2+k^2)^n}&=\int_{-\infty}^\infty\frac1{\Gamma(n)}\int_0^\infty t^{n-1}e^{-t(x^2+k^2)}\mathrm dt\mathrm dx \\ &=\frac1{\Gamma(n)}\int_0^\infty t^{n-1}e^{-tk^2}\left(\int_{-\infty}^\infty e^{-tx^2}\mathrm dx\right)\mathrm dt\\ &=\frac1{\Gamma(n)}\int_0^\infty t^{n-1}e^{-tk^2}\sqrt{\frac\pi t}\mathrm dt\\ &=\frac{\sqrt\pi}{\Gamma(n)}\int_0^\infty t^{n-3/2}e^{-tk^2}\mathrm dt \\ &=\frac{\sqrt\pi\Gamma\left(n-\frac12\right)}{\Gamma(n)k^{2n-1}}. \end{align} $$

We used the fact that

$$\int_0^\infty t^{n-1}e^{-At}\mathrm dt=\frac{\Gamma(n)}{A^n}\iff\frac{1}{A^n}=\frac{1}{\Gamma(n)}\int_0^\infty t^{n-1}e^{-At}\mathrm dt.$$

Clever use of a Laplace transform. – Sean Roberson Mar 29 '25 at 00:08 — Sean Roberson, Mar 29 '25 at 00:08

Any general formulas for evaluating integrals of the form $ \int_{-\infty}^{\infty} \frac{1}{(x^2+k^2)^n} \,dx $

1 Answers1