Question

Find a closed form for the family of integrals,
\[\mathcal{I}_n=\int_{\mathbb{R}}\frac{e^{-x^2}}{x^2+n}\,\mathrm{d}x\]where \(n\in\mathbb{N}\).

Answer 1

\[\mathcal{I}_n(1)=\int_{\mathbb{R}}\frac{e^{-x^2}}{x^2+n}\,\mathrm{d}x\]
\[\mathcal{J}_n(t)=\int_{\mathbb{R}}\frac{e^{-(x^2+n)t}}{x^2+n}\,\mathrm{d}x\]

I'm like falling asleep awake over 24 hours but like this was gonna be my attempt so hopefully it differentiates and integrates back well

Answer 2

I have an approach in mind using a different parameterization and the Laplace transform, but I'm stuck on one part (which I admit I haven't spent much time working out):
\[\begin{align*}\mathcal{I}_{a,n}=\int_{-\infty}^\infty\frac{e^{-ax^2}}{x^2+n}\,\mathrm{d}x\implies \mathcal{L}\left\{\mathcal{I}_{a,n}\right\}&=\int_{-\infty}^\infty\frac{1}{x^2+n}\mathcal{L}\left\{e^{-ax^2}\right\}\,\mathrm{d}x\\[1ex]
&=\frac{\pi}{\displaystyle s\sqrt n+n\sqrt s}\end{align*}\]
I'm considering writing the denominator as \(\sqrt n\left(\sqrt s+\dfrac{\sqrt n}{2}\right)^2-\dfrac{n^{3/2}}{4}\), but that doesn't make things much easier. It does kinda resemble the Laplace transform of a hyperbolic sine function, but not quite, thanks to the \(\sqrt s\).

The inverse transform might be doable using the contour integral definition, but I'm not familiar enough with branch points to either know how to compute that integral or even know whether or not they're relevant.

Answer 3

It would be nice if there was a known inverse Laplace transform formula for \(f(t)\) given \(F(\sqrt s)\).

Answer 4

Gaussian integral fact:\[\int_{-\infty}^\infty e^{-tx^2}dx = \sqrt{\pi} t^{-1/2}\]
multiply through by \(e^{-tn}\)
\[\int_{-\infty}^\infty e^{-(x^2+n)t}dx = \sqrt{\pi} t^{-1/2}e^{-nt}\]
integrate wtf to t

\[\int \int_{-\infty}^\infty e^{-(x^2+n)t}dxdt = \int \sqrt{\pi} t^{-1/2}e^{-nt}dt \]
independent so reverse order of integration on the left side easy:

\[\int_{-\infty}^\infty \int e^{-(x^2+n)t}dtdx = \int \sqrt{\pi} t^{-1/2}e^{-nt}dt \]

\[-\int_{-\infty}^\infty \frac{ e^{-(x^2+n)t}}{x^2+n}dx = \int \sqrt{\pi} t^{-1/2}e^{-nt}dt \]
That looks familiar but how do we deal with the RHS? It's a substitution away from a gamma, but I think that's fine as long as we throw in an integration constant to undo it maybe?

\[-\int_{-\infty}^\infty \frac{ e^{-(x^2+n)t}}{x^2+n}dx = \sqrt{\pi} \int t^{-1/2}e^{-nt}dt \]

Anyways at t=1 you can pull out the \(e^{-n}\) factor we threw in earlier to get your integral as it should be. I don't know if this is what you're doing or not or different.

Answer 5

That's not what I'm doing (I think). I don't think we can get any information containing the \(\Gamma\) function from the RHS of your equations since it's defined by a definite integral.

Answer 6

I realized I neglected showing the work for the expression I arrived at in my first post. I'll include it for the sake of completeness.

Parameterizing the integral as
\[\mathcal{I}_{a,n}=\int_{-\infty}^\infty \frac{e^{-ax^2}}{x^2+n}\,\mathrm{d}x\]and taking the Laplace transform, we have
\[\begin{align*}
\mathcal{L}\left\{\mathcal{I}_{a,n}\right\}&=\int_{-\infty}^\infty \frac{1}{x^2+n}\mathcal{L}_{a,s}\left\{e^{-ax^2}\right\}\,\mathrm{d}x\\[1ex]

&=\int_{-\infty}^\infty\frac{1}{(x^2+n)(x^2+s)}\,\mathrm{d}x\\[1ex]

&=\frac{1}{s-n}\int_{-\infty}^\infty\left(\frac{1}{x^2+n}-\frac{1}{x^2+s}\right)\,\mathrm{d}x\\[1ex]

&=\frac{1}{s-n}\left(\frac{\pi}{\sqrt n}-\frac{\pi}{\sqrt s}\right)\\[1ex]
&=\frac{\pi}{s-n}\left(\frac{\sqrt s-\sqrt n}{\sqrt {ns}}\right)\\[1ex]
&=\frac{\pi}{\sqrt{ns}\left(\sqrt s+\sqrt n\right)}\\[1ex]
&=\frac{\pi}{s\sqrt n+n\sqrt s}
\end{align*}\]
(where \(\mathcal{L}_{a,s}\) denotes a transform from the \(a\) domain to the \(s\) domain.)

Answer 7

Integrating by parts might be another possible route to take. With
\[\begin{matrix}u=e^{-x^2}&&&\mathrm{d}v=\dfrac{\mathrm{d}x}{x^2+n}\\[1ex]
\mathrm{d}u=-2xe^{-x^2}\,\mathrm{d}x&&&v=\dfrac{1}{\sqrt n}\arctan\dfrac{x}{\sqrt n}\end{matrix}\]we get
\[\begin{align*}\mathcal{I}_n&=\frac{1}{\sqrt n}\bigg[e^{-x^2}\arctan\frac{x}{\sqrt n}\bigg]_{-\infty}^\infty+\frac{2}{\sqrt n}\int_{-\infty}^\infty xe^{-x^2}\arctan\frac{x}{\sqrt n}\,\mathrm{d}x\\[1ex]
&=\frac{2}{\sqrt n}\int_{-\infty}^\infty xe^{-x^2}\arctan\frac{x}{\sqrt n}\,\mathrm{d}x\\[1ex]
&=\frac{4}{\sqrt n}\int_0^\infty xe^{-x^2}\arctan\frac{x}{\sqrt n}\,\mathrm{d}x\\[1ex]
&=2\sqrt n\int_0^\infty e^{-nt}\arctan\sqrt t\,\mathrm{d}t\end{align*}\]where \(\sqrt t=\dfrac{x}{\sqrt n}\).

Easier? Maybe...

Answer 8

Another attempt that uses a different IBP setup:
\[\begin{matrix}
u=\dfrac{e^{-x^2}}{x^2+n}&&&\mathrm{d}v=\mathrm{d}x\\[1ex]
\mathrm{d}u=-\dfrac{2xe^{-x^2}\left(x^2+n+1\right)}{\left(x^2+n\right)^2}\,\mathrm{d}x&&&v=x
\end{matrix}\]which gives
\[\mathcal{I}_n=2\int_{-\infty}^\infty \frac{x^2e^{-x^2}\left(x^2+n+1\right)}{\left(x^2+n\right)^2}\,\mathrm{d}x\]and splitting into partial fractions yields
\[\begin{align*}\mathcal{I}_n&=2\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x-2n\int_{-\infty}^\infty \frac{e^{-x^2}}{\left(x^2+n\right)^2}-2(n-1)\int_{-\infty}^\infty \frac{e^{-x^2}}{x^2+n}\,\mathrm{d}x\\[1ex]
&=2\sqrt\pi-2n\int_{-\infty}^\infty \frac{e^{-x^2}}{\left(x^2+n\right)^2}-2(n-1)\mathcal{I}_n\\[1ex]
(2n-1)\mathcal{I}_n&=2\sqrt\pi-2n\int_{-\infty}^\infty \frac{e^{-x^2}}{\left(x^2+n\right)^2}\end{align*}\]It makes me wonder about this even more general family of integrals: for \(p\in\mathbb{N}\),
\[\mathcal{I}_{n,p}=\int_\mathbb{R}\frac{e^{-x^2}}{\left(x^2+n\right)^p}\,\mathrm{d}x\]