Difficult infinite integral involving a Gaussian, Bessel function and complex singularities

I've come across the following integral in my work. $$\intop_{0}^{\infty}dk\, e^{-ak^{2}}J_{0}\left(bk\right)\frac{k^{3}}{c^{2}+k^{4}} $$

Where $a$,$b$,$c$ are all positive.

I've seen and evaluate similar integrals without the denominator using resources like Watson's Theory of Bessel Functions, but I've had no luck finding anything resembling this integral. It would get me a very awesome result if I were to evaluate this. Does anyone have any ideas on how to approach it?

edit: Some of my attempts so far include:

1)Using "Infinite integrals involving Bessel functions by an improved approach of contour integration and the residue theorem" by Qiong-Gui Lin. This (like other residue theorem approaches) doesn't seem to work since the gaussian blows up on the imaginary axis.

2)I recall seeing expressions like $Z_u(ax)X_v(b\sqrt x)$ where Z,X are some variation of bessel functions in Gradshteyn, Ryzhik. This inspired me to write $e^{-ax^2}=\sum_{k=-\infty}^{\infty}I_{k}(-ax^{2})$ and substitute $t=x^2$, then integrate term by term. This hasn't gotten me anywhere either.

Solution 1:

I shall describe an expansion for this integral $\mathcal{I}(a,b,c)$ in powers of $c^{-2}$ . To do so I will make a few changes of parameters first. Observe that the substitution $x=bk$ yields $$ \mathcal{I}(a,b,c) =\int_0^\infty dx \,e^{-a x^2/b^2} J_0(x)\frac{x^3}{b^{4} c^{2}+x^4}.$$ Defining $t=b^2/4a$, $\epsilon=1/b^4 c^2,$ and $I(\alpha,\epsilon) = b^4 c^2 \mathcal{I}(a,b,c)$, we have $$I(t,\epsilon) = \int_0^\infty dx\, e^{-x^2/4t} J_0(x)\frac{x^3}{1+\epsilon x^4}. \tag{1}$$

With this form in hand, we expand in powers of $\epsilon\sim c^{-2}$ to obtain $$I(t,\epsilon)=\sum_{k=0}^\infty (-\epsilon)^k \int_0^\infty dx\, x^{3+4k} e^{-x^2/4t}J_0(x).$$ The resulting term-by-term integration may be treated using formula 6.631.1 of Gradshteyn and Ryzhik (for reference, this is with $(\mu,\nu,\alpha,\beta)=(3+4k,0,1/4t,1))$: $$\int_0^{\infty} dx\,x^{3+4k} e^{-x^2/4t}J_0(x) = \frac{1}{2}(4t)^{2k+2}(2k+1)!\,_1F_1(2k+2;1;-t)$$ where $ _1F_1(a;1;t)$ is Kummer's confluent hypergeometric series. This satisfies Kummer's transformation, allowing us to write \begin{align} _1F_1(2k+2;1;-t) &=e^{-t}\,_1F_1(-1-2k;1;t)\\ &=e^{-t} \sum_{j=0}^{2k+1} \frac{(-1-2k)_j}{(j!)^2}t^j=\sum_{j=0}^{2k+1}\binom{2k+1}{j}\frac{t^j}{j!}e^{-t} \tag{2} \end{align} where the summation is terminated by the negative argument of the rising factorial $(x)_n$.

Recalling that the definition of the $n$th Laguerre polynomial is $L_n(x)=\sum_{k=0}^n\dfrac{(-x)^k}{k!}$, we may write equation $(2)$ as $e^{-t} L_{2k+1}(t)$. Hence we may express equation $(1)$ as \begin{align} I(t,\epsilon) &=\sum_{k=0}^\infty (-\epsilon)^k \frac{1}{2}(4t)^{2k+2}(2k+1)! \, e^{-t} L_{2k+1}(t)\\ &=2t e^{-t} \sum_{m\text{ odd}}^\infty (-\epsilon)^{\frac{m-1}{2}} (4t)^m m! \, L_m(t)\ \end{align} [to be continued]

Solution 2:

To find the behavior when $c \to 0$ we'll split the integral into the two pieces

$$ \int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} = \left( \int_{0}^{\sqrt{c}} + \int_{\sqrt{c}}^{\infty} \right) dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4}. \tag{1} $$

The first piece can be expanded in powers of $c$ by writing

$$ e^{-ak^2} J_0(bk) = \sum_{j=0}^{\infty} \alpha_j k^{2j}, $$

then substituting this in to find that

$$ \begin{align} &\int_{0}^{\sqrt{c}} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\ &\qquad = \sum_{j=0}^{\infty} \alpha_j \int_0^\sqrt{c} \frac{k^{3+2j}}{c^2+k^4}\,dk \\ &\qquad = \sum_{j=0}^{\infty} \left(\alpha_j \int_0^1 dx\, \frac{x^{3+2j}}{1+x^4} \right) c^j \\ &\qquad = \frac{1}{4}\log 2 - \frac{(4-\pi)(4a-b^2)}{32} c + \frac{(1-\log 2)(32 a^2+16 a b^2+b^4)}{256} c^2 + \cdots, \end{align} $$ $$ \tag{2} $$

so it stands to reason that the leading-order asymptotic behavior of the integral comes from the second piece. Let's use the fact that

$$ \frac{k^3}{c^2+k^4} = \frac{1}{k} - \frac{c^2}{k(c^2+k^4)} $$

to write it as

$$ \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} = \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} - c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} $$ $$ \tag{3} $$

Fix $0 < \epsilon < 1/2$ and split the second integral like

$$ c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = c^2 \left(\int_{\sqrt{c}}^{\Large c^\epsilon} + \int_{\Large c^\epsilon}^{\infty} \right) dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)}. $$ $$ \tag{4} $$

The tail term is bounded by

$$ \left| c^2 \int_{\Large c^\epsilon}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} \right| < c^2 \int_{\Large c^\epsilon}^\infty \frac{dk}{k (c^2+k^4)} = \frac{1}{4} \log(1 + c^{2-4\epsilon}) \tag{5} $$

and the first can be expanded as before;

$$ c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = c^2 \sum_{j=0}^{\infty} \alpha_j \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, \frac{k^{2j}}{k(c^2+k^4)}. $$

In fact we'll show that all we can use are the first two terms of this expansion, so for now we'll just write

$$ \begin{align} &c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} \\ &\qquad = c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} - \left(a+\frac{b^2}{4}\right) c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k}{c^2+k^4} + O\left(c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k^3}{c^2+k^4}\right). \end{align} $$ $$ \tag{6} $$

We must now estimate these new integrals. For the first we make the change of variables $k = \sqrt{c} x$ to get

$$ c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} = \int_1^{\Large c^{\epsilon - 1/2}} \frac{dx}{x (1+x^4)} = \int_1^\infty \frac{dx}{x (1+x^4)} - \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x (1+x^4)}. $$

Of course

$$ \int_1^\infty \frac{dx}{x (1+x^4)} = \frac{1}{4}\log 2 $$

and

$$ 0 < \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x (1+x^4)} < \int_{\Large c^{\epsilon-1/2}}^\infty \frac{dx}{x^5} = \frac{1}{4} c^{2-4\epsilon}, $$

so we just end up with

$$ c^2 \int_\sqrt{c}^{\Large c^\epsilon} \frac{dk}{k (c^2+k^4)} = \frac{1}{4}\log 2 + O(c^{2-4\epsilon}). \tag{7} $$

An identical argument applied to the second integral yields

$$ c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k}{c^2+k^4} = \frac{\pi}{8} c + O(c^{2-4\epsilon}). \tag{8} $$

For the last integral we only need the blunt estimate

$$ 0 < c^2 \int_\sqrt{c}^{\Large c^\epsilon} dk\, \frac{k^3}{c^2+k^4} = c^2 \int_1^{\Large c^{\epsilon-1/2}} \frac{x^3}{1+x^4} < c^2 \log c^{\epsilon-1/2} < c^{2-4\epsilon} $$

for $c$ small enough. By combining this with $(7)$ and $(6)$ in $(5)$ we get

$$ c^2 \int_{\sqrt{c}}^{\Large c^\epsilon} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = \frac{1}{4}\log 2 - \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-4\epsilon}), \tag{9} $$

and this, combined with $(5)$ in $(4)$, yields

$$ c^2 \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{1}{k(c^2+k^4)} = \frac{1}{4}\log 2 - \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-4\epsilon}). \tag{10} $$

Thus $(3)$ becomes

$$ \begin{align} &\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\ &\qquad = \int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} - \frac{1}{4}\log 2 + \frac{\pi}{8} \left(a+\frac{b^2}{4}\right) c + O(c^{2-\epsilon}) \tag{11} \end{align} $$

as $c \to 0^+$ for any fixed $\epsilon > 0$. Finally, we can write the integral here as

$$ \begin{align} &\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} \\ &\qquad = \int_{\sqrt{c}}^{\infty} dk\, e^{-k} k^{-1} + \int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\ &\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + \int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1}, \end{align} $$

where $\operatorname{Ei}$ is the exponential integral. Now

$$ \begin{align} &\int_{\sqrt{c}}^{\infty} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\ &\qquad = \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} - \int_0^\sqrt{c} dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} \\ &\qquad = f(a,b) - \sum_{j=0}^{2} \beta_j \int_0^\sqrt{c} dk\, k^j + O(c^2) \\ &\qquad = f(a,b) - \sqrt{c} + \frac{1}{2}\left(a + \frac{b^2}{4} + \frac{1}{2}\right) c - \frac{1}{18} c^{3/2} + O(c^2), \end{align} $$

where

$$ f(a,b) := \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} $$

and the coefficients $\beta_j$ are defined by

$$ \begin{align} \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1} &= \sum_{j=0}^{\infty} \beta_j k^j \\ &= 1 - \left(a + \frac{b^2}{4} + \frac{1}{2}\right) k + \frac{1}{6} k^2 + \cdots, \end{align} $$

so that

$$ \begin{align} &\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) k^{-1} \\ &\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \sqrt{c} + \frac{1}{2}\left(a + \frac{b^2}{4} + \frac{1}{2}\right) c - \frac{1}{18} c^{3/2} + O(c^2). \end{align} $$

Substituting this into $(11)$ thus yields

$$ \begin{align} &\int_{\sqrt{c}}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\ &\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \frac{1}{4}\log 2 - \sqrt{c} + \left( \frac{1}{4} + \frac{\pi+4}{8}a + \frac{\pi+4}{32} b^2\right) c \\ &\qquad \qquad - \frac{1}{18}c^{3/2} + O(c^{2-\epsilon}), \end{align} $$ $$ \tag{12} $$

and, at last, combining this with $(2)$ in $(1)$ grants us

$$ \begin{align} &\int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\ &\qquad = -\operatorname{Ei}\left(-\sqrt{c}\right) + f(a,b) - \sqrt{c} + \frac{1 + \pi a + b^2}{4} c - \frac{1}{18} c^{3/2} + O(c^{2-\epsilon}). \tag{13} \end{align} $$

It is known (see wikipedia) that

$$ \operatorname{Ei}(z) = \log|z| + \gamma + x + \frac{1}{4}x^2 + \frac{1}{18}x^3 + O(x^4) $$

as $x \to 0$, so in our case we have

$$ -\operatorname{Ei}\left(-\sqrt{c}\right) = \frac{1}{2} \log \frac{1}{c} - \gamma + \sqrt{c} - \frac{1}{4} c + \frac{1}{18} c^{3/2} + O(c^2), $$

and so we arrive at the asymptotic

$$ \begin{align} &\int_{0}^{\infty} dk\, e^{-ak^2} J_0(bk) \frac{k^3}{c^2+k^4} \\ &\qquad = \frac{1}{2} \log \frac{1}{|c|} + f(a,b) - \gamma + \frac{\pi a+b^2}{4} |c| + O(|c|^{2-\epsilon}) \tag{14} \end{align} $$ as $c \to 0$ for any fixed $\epsilon > 0$, where $$ f(a,b) = \int_0^\infty dk\, \Bigl( e^{-ak^2} J_0(bk) - e^{-k} \Bigr) k^{-1}. $$

Surely the $O(|c|^{2-\epsilon})$ term may be replaced with $\Theta(c^2 \log |c|)$ with a little more work.

Here's a log-log plot to illustrate the asymptotic with $a=b=1$ over the range $c \in (2\cdot 10^{-4},10^{-2})$. The black points are numerical evaluations of the given integral and the blue curve is

$$ \frac{1}{2} \log \frac{1}{c} + f(1,1) - \gamma. $$