Integrating $\int_{0}^{\infty} x^{-k} e^{-a(x+d)^2} dx$ where $k \in \mathbb{Z}_{+}$

Question

I want to solve the following integral:

$$I = \int_{0}^{\infty} x^{-k} e^{-a(x+d)^2} dx$$

where $k \in \mathbb{Z}_{+}$, and $a, d \in \mathbb{R}_{++}$.

When $k =0$, the integral is a standard Gaussian integral. However, when $k > 0$, the problem is not so easy. We see that when $x \rightarrow 0$, the integrand blows up.

We cannot set $I = \underset{\epsilon \rightarrow 0^{+}}{\text{lim}} \int_{0}^{\infty} x^{-k} e^{-a(x+d)^2} dx $ and then expand $x^{-k}$ into its Maclaurian series because we are integrating over all $x > 0$ and the series converges only within a finite region. Similarly, we cannot put a contour that goes around the singularity because the pole is of higher order. Also, solving as if $k < 0$ and then using analytic continuation of the Gamma function won't work, because $k$ is an integer.

Any ideas? Or does this integral simply not exist?

Thanks

Answer 1

$$I=\int_0^\infty x^{-k}e^{-[\sqrt{a}(x+d)]^2}dx$$ $u=\sqrt{a}(x+d)\Rightarrow dx=du/\sqrt{a}$ so we have: $$I=\frac{1}{\sqrt{a}}\int_{\sqrt{a}d}^\infty\left(\frac{u}{\sqrt{a}}-d\right)^{-k}e^{-u^2}du$$ $$=a^{(k-1)/2}\int_{\sqrt{a}d}^\infty(u-\sqrt{a}d)^{-k}e^{-u^2}du$$ now try and expand $(u-\sqrt{a}d)^{-k}$ to get this as a series of integrals and then use the incomplete gamma function

Answer 2

This integral does not exist. Consider the case when $k=1$. (I think other cases reduce to this via integration by parts.) We know $\int_0^n\frac{dx}{x}$ diverges. But our integral $\int_0^n\frac{e^{-a(x+d)^2}dx}{x}\ge\int_0^n\frac{mdx}{x}$ where $m$ is the minimum of $e^{-a(x-d)^2}$ on $[0,n]$.