# Integral involving a power function and $\exp(-a/x^2-b x^2)$.

This is a generalization of Double integral containing $e^{(b+ic)/z^2}$ . Let us take $$a >0$$, $$b>0$$. The following results are well known: $$\begin{eqnarray} \int\limits_0^\infty e^{-\frac{a}{x^2} - b x^2} dx &=& \frac{\sqrt{\pi}}{2} \frac{e^{-2 \sqrt{a b}}}{\sqrt{b}} \quad (i)\\ \int\limits_0^\infty x e^{-\frac{a}{x^2} - b x^2} dx &=& \sqrt{\frac{a}{b}} K_1(2 \sqrt{2 a b} ) \quad (ii) \end{eqnarray}$$ They come from integral representations of Bessel functions. Now it would be natural to take a number $$\theta \in (0,1)$$ and ask for a "fractional moment" of the integrand. As a matter of fact the following result holds:

$$\begin{eqnarray} \int\limits_0^\infty x^{2 \theta} e^{-\frac{a}{x^2} - b x^2} dx = \frac{\sqrt{\pi}}{2 (-\theta)!} b^{-\frac{1}{2}-\theta} \int\limits_0^\infty e^{-2 \sqrt{a b} \cosh(\phi)} \frac{\left( -1+ 2\theta +2 \sqrt{a b} \cosh(\phi)\right)}{ [\sinh(\phi)]^{-1+2\theta} }d\phi \end{eqnarray}$$

In[2167]:= {a, b, th} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
NIntegrate[x^(2 th) Exp[-a/x^2 - b x^2], {x, 0, Infinity}]

t = Range[-1/2, 1/2, 0.001];
(*ll0=Transpose[{t, NIntegrate[1/(#+xi)^th(\[ExponentialE]^(-2 Sqrt[a \
(b+xi)]) Sqrt[\[Pi]])/(2 Sqrt[b+xi]),{xi,-#,Infinity}]&/@t}];*)
(*ll=Transpose[{t,Sqrt[\[Pi]] NIntegrate[\[ExponentialE]^(-2 Sqrt[a \
]u)/(u^2-(b-#))^th ,{u,Sqrt[b-#],Infinity}]&/@t}];*)
ll = Transpose[{t,
Sqrt[\[Pi]] (b - #)^(1/2 - th)
NIntegrate[
E^(-2 Sqrt[a ] Sqrt[(b - #)] Cosh[phi]) Sinh[phi]^(
1 - 2 th), {phi, 0, Infinity}] & /@ t}];
f = Interpolation[ll]
1/(-th)! f'[0]

1/(-th)! Sqrt[\[Pi]] 1/2 b^(-(1/2) - th)
NIntegrate[
E^(-2 Sqrt[a] Sqrt[b]
Cosh[phi]) (-1 + 2 th + 2 Sqrt[a] Sqrt[b] Cosh[phi]) Sinh[phi]^(
1 - 2 th), {phi, 0, Infinity}]
1/(-th)! Sqrt[\[Pi]] 1/2 b^(-(1/2) - th)
NIntegrate[(
2^(-1 + 2 th) E^(-Sqrt[a] Sqrt[b] (1/z + z)) (-(1/z) + z)^(
1 - 2 th) (-1 + 2 th + Sqrt[a] Sqrt[b] (1/z + z)))/
z , {z, 1, Infinity}]
1/(-th)! Sqrt[\[Pi]] 2^(-2 + 2 th) b^(-(1/2) - th)
NIntegrate[(
E^(-Sqrt[a] Sqrt[b] (1/z + z)) (1/z - z)^(
1 - 2 th) (-1 + 2 th + Sqrt[a] Sqrt[b] (1/z + z)))/z , {z, 0, 1}]
1/(-th)! Sqrt[\[Pi]] 2^(-2 + 2 th) b^(-(1/2) - th)
NIntegrate[
E^(-Sqrt[a] Sqrt[b]
u) (-1 + 2 th + Sqrt[a] Sqrt[b] u) (-4 + u^2)^-th, {u, 2,
Infinity}]
1/(-th)! Sqrt[\[Pi]] 2^(-2 + 2 th) b^(-(1/2) - th)
NIntegrate[(
E^(-Sqrt[a] Sqrt[b] Sqrt[4 + v^2])
v (v^2)^-th (-1 + 2 th + Sqrt[a] Sqrt[b] Sqrt[4 + v^2]))/Sqrt[
4 + v^2], {v, 0, Infinity}]
1/(-th)! Sqrt[\[Pi]] 2^-1 b^(-(1/2) - th)
NIntegrate[
E^(-2 Sqrt[a] Sqrt[b]
Cosh[phi]) (-1 + 2 th + 2 Sqrt[a] Sqrt[b] Cosh[phi])  Sinh[phi]^(
1 - 2 th), {phi, 0, Infinity}]

Out[2168]= 8.86647

Out[2172]= 8.8673 + 0. I

Out[2173]= 8.86647

Out[2174]= 8.86647

Out[2175]= 8.86647

Out[2176]= 8.86647

Out[2177]= 8.86647

Out[2178]= 8.86647

We have obtained this result by fractionally differentiating $$(i)$$ with respect to $$b$$. Now having said all this can we simplify the right hand side any further , meaning express it in terms of Bessel functions itself ? Another question would be what could this result be useful for ? Has this quantity been analyzed already before and if yes in what context ?

• Why not to use the integral representation of the modified Bessel function to obtain directly $$I=\left( \frac{a}{b} \right)^{\theta/2+1/4}K_{\theta+1/2}\left( \sqrt{2ab} \right)$$ – Paul Enta Mar 29 at 23:09
• @Paul Enta Thank you for that. This implies that $\int\limits_1^\infty \frac{\left(t-\frac{1}{t}\right)^{1-2 \text{th}} e^{-\sqrt{a} \sqrt{b} \left(t+\frac{1}{t}\right)} \left(\sqrt{a} \sqrt{b} \left(t+\frac{1}{t}\right)+2 \text{th}-1\right)}{t} dt=\frac{4^{1-\text{th}} (-\text{th})! (a b)^{\frac{1}{4} (2 \text{th}+1)} K_{-\text{th}-\frac{1}{2}}\left(2 \sqrt{a b}\right)}{\sqrt{\pi }}$ for $a>0$,$b>0$ and $\theta \in (0,1)$. – Przemo Apr 1 at 9:51
• You are welcome. Your new identity can be silightly simplified as it depends on the parameter $\sqrt({ab}$. – Paul Enta Apr 1 at 12:45