I wish to compute the following function
$$\forall\ \mathbf{b}\in \mathbb{R}^3,\quad V(\mathbf{b}) := \int_{\mathbb{R}^{3}} \frac{ e^{-\alpha \mathbf{k}^2 + \mathbf{b}\cdot \mathbf{k} } }{ \sqrt{\mathbf{k}^2+m^2}}\, d^{3} \mathbf{k} \quad \text{where}\enspace \mathbf{k}^2:= \lVert \mathbf{k}\rVert^2 $$
I thought of looking for an ODE it would satisfy, but to no avail up to now. A change to spherical coordinates with "$\mathbf{b}$" as the $z$ axis yields
$$\begin{aligned}
& \iiint \frac{ e^{-\alpha\hspace{.5pt} k^2\hspace{.5pt} + \hspace{.5pt}b\, k\hspace{.5pt} \cos \theta } }{ \sqrt{k^2+m^2} }\, k^2 \hspace{.7pt} \sin \theta\, d k\, d\theta\, d\varphi = 2\hspace{.7pt}\pi \int_0^{+\infty} \frac{ k^2\, e^{-\alpha\hspace{.5pt} k^2\hspace{.5pt} } }{ \sqrt{k^2+m^2} }\, \left[\frac{e^{\hspace{.5pt} b\, k\hspace{.5pt} \cos \theta}}{- \hspace{.5pt} b\, k}\right]_0^{\pi}\, dk \quad \text{where}\enspace k:= \lVert \mathbf{k}\rVert,\ b:= \lVert \mathbf{b}\rVert \\
&= \frac{2\hspace{.7pt} \pi}{b} \int_0^{+\infty} \frac{ k\left( e^{-\alpha\hspace{.5pt} k^2\hspace{.5pt} + b\hspace{.3pt} k } – e^{-\alpha\hspace{.5pt} k^2\hspace{.5pt} – b\hspace{.3pt} k } \right)}{ \sqrt{k^2+m^2}}\, dk
\end{aligned}$$
Inserting then the following definition of Hermite polynomials:
$\quad \displaystyle e^{2\hspace{.3pt}x\hspace{.3pt}t – t^2} = \sum_{n=0}^{+\infty} H_n(x)\, \frac{t^n}{n!} $ with $t:= \sqrt{\alpha}\, k,\ x := \frac{b}{2\sqrt{\alpha}} $
$$ \begin{aligned}
V(b) &= \frac{2\hspace{.7pt} \pi}{b} \int_0^{+\infty} \frac{k}{ \sqrt{k^2+m^2}} \sum_{n=0}^{+\infty} \left( H_n\left(\frac{b}{2\sqrt{\alpha}}\right) – H_n\left(\frac{-\, b}{2\sqrt{\alpha}}\right) \right) \frac{\left(\sqrt{\alpha}\, k\right)^n}{n!} \, dk \\
&= \frac{2\hspace{.7pt} \pi}{b} \sum_{n=0}^{+\infty} H_n\left(\frac{b}{2\sqrt{\alpha}}\right) \big( 1 + (-1)^{n+1}\big) \frac{\left(\sqrt{\alpha}\right)^n}{n!} \int_0^{+\infty} \frac{k^{n+1}}{\sqrt{k^2+m^2}} \, dk
\end{aligned}$$
The last integral formally looks like a Beta function
after the following change of variable $\genfrac{[}{]}{0pt}{0}{l:= \frac{k^2}{m^2}}{dl = \frac{2}{m^2}\, k\, dk}$
$$ \int_0^{+\infty} \frac{k^{n+1}}{m \sqrt{k^2/m^2 + 1}} \, dk = \frac{m}{2}\int_0^{+\infty} \frac{m^n\, l^{\frac{n}{2}}}{\sqrt{l+1}} \, dl = \frac{m^{n+1}}{2} B\left(\frac{n}{2}+ 1 , -\frac{n+1}{2}\right)$$
Indeed, the integral is bluntly divergent and one should not have permuted sum and integral…
Question: can $V$ be expressed in terms of special functions? or as a power series in $b$? (in fact I did found one not for $\lVert \mathbf{b}\rVert$, but as a function of each component $b_1, b_2, b_3$ and it is desperately involved… so I still keep on looking for something more reasonnable…)
I also tried integration by parts on
$$V(b) = \frac{2\hspace{.7pt} \pi}{b} \int_0^{+\infty} \frac{ k }{ \sqrt{k^2+m^2}} \times e^{-\alpha\hspace{.5pt} k^2 }\, \sinh(b\hspace{.3pt} k)\, dk$$
and complex integration…
EDIT: I meant PDE above… in fact I've just found an ODE for
$$U(b):= \int_0^{+\infty} \frac{ e^{-\alpha\hspace{.5pt} k^2 }\, \cosh(b\hspace{.3pt} k) }{ \sqrt{k^2+m^2}}\, dk$$
Indeed
$$ \begin{aligned}
U'(b) & = \int_0^{+\infty} \frac{ k\, e^{-\alpha\hspace{.5pt} k^2 }\, \sinh(b\hspace{.3pt} k) }{ \sqrt{k^2+m^2}}\, dk\\
(\text{Int. by part}) &= \left[ \sqrt{k^2+m^2}\, e^{-\alpha\hspace{.5pt} k^2 }\, \sinh(b\hspace{.3pt} k) \right]_0^{+\infty} – \int_0^{+\infty} \sqrt{k^2+m^2} \left[ (- 2 \alpha k) e^{-\alpha\hspace{.5pt} k^2 }\, \sinh(b\hspace{.3pt} k) +b\, e^{-\alpha\hspace{.5pt} k^2 }\, \cosh(b\hspace{.3pt} k) \right]\, dk
\end{aligned}$$
The "boundary term" in the integral by part vanishes, and the other can be expressed in terms of the following:
$$ U''(b) = \int_0^{+\infty} \frac{k^2 e^{-\alpha\hspace{.5pt} k^2 }\, \cosh(b\hspace{.3pt} k) }{ \sqrt{k^2+m^2}}\, dk = \int_0^{+\infty} \frac{\big( k^2 +m^2 – m^2\big) e^{-\alpha\hspace{.5pt} k^2 }\, \cosh(b\hspace{.3pt} k) }{ \sqrt{k^2+m^2}}\, dk $$
Then consider also $U'''(b) +m^2 U'(b)$. Shake everything, and you should find a linear ODE… with non constant coeff, so there will still be work…
Best Answer
Considering $$I = \int_0^{\infty} \frac{ k }{ \sqrt{k^2+m^2}} \, e^{-\alpha\,k^2 }\, \sinh(b\, k)\, dk$$ Using the Taylor series of the hyperbolic sine function $$I=\sum_{n=0}^\infty \frac{b^{2 n+1}}{(2 n+1)!} \int_0^{\infty}\frac{ k^{2n+2} }{ \sqrt{k^2+m^2}} \, e^{-\alpha\,k^2 }\,dk$$ $$I=\sum_{n=0}^\infty \frac{b^{2 n+1}}{(2 n+1)!}\,J_n$$ $$J_n= \int_0^{\infty}\frac{ k^{2n+2} }{ \sqrt{k^2+m^2}} \, e^{-\alpha\,k^2 }\,dk$$ $$J_n=\sqrt{\pi }\,\frac{ (2 n+1)\text{!!}}{2^{n+2}\, \alpha ^{n+1} }\,\,U\left(\frac{1}{2},-n,\alpha m^2\right)$$ where $U(.)$ is Tricomi's confluent hypergeometric function.
Defining $$A_n=\frac{b^{2 n+1}}{(2 n+1)!}\,J_n$$ $$\frac{A_{n+1}}{A_n}=\frac{b^2}{4 \alpha (n+1)}\,\frac{U\left(\frac{1}{2},-(n+1),m^2 \alpha \right)}{U\left(\frac{1}{2},-n,m^2 \alpha \right)}$$
$$\frac{b^2}{2 \alpha (2 n+3)}\leq \frac{A_{n+1}}{A_n} <\frac{b^2}{2 \alpha (2n+2)}$$