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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.

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  • $\begingroup$ One obvious point is that at least one of the three parameters is superfluous i.e. can be absorbed into the definition of the other by appropriate rescaling. (Since the Bessel function is the most opaque, I'd probably use this to set $b\to 1$.) $\endgroup$ – Semiclassical Jul 16 '14 at 1:22
  • $\begingroup$ Also, do you have sense of how large (relatively speaking) these parameters will be? For example, if $c^2$ is typically small, then one sensible tack would be to expand the denominator in a power series in $c^2$ and integrate term-by-term. But that's unwise if you want the large $c$ limit... $\endgroup$ – Semiclassical Jul 16 '14 at 1:39
  • $\begingroup$ There is not much to be said about $a$ or $b$, but I am certainly interested in large and small limits for $c$. I'll try out expanding like you've said for small $c$. $\endgroup$ – David T Jul 16 '14 at 1:59
  • $\begingroup$ For large $c$, you'd just expand in powers of $1/c^2$ instead. $\endgroup$ – Semiclassical Jul 16 '14 at 2:01
  • $\begingroup$ What kinds of further results are you hoping for? $\endgroup$ – Semiclassical Jul 24 '14 at 23:39
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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]

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  • $\begingroup$ Nice work... It seems like we get $I(a,b,c)\sim c^{-2}$ for very large $c$. That is a good result. $\endgroup$ – David T Jul 16 '14 at 6:53
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    $\begingroup$ Just noticed a small mistake.. I think $$\int_0^{\infty} dx\,x^{3+4k} e^{-x^2/4t}J_0(x) = \frac{1}{2}(8t)^{2k+2}(2k+1)!\,_1F_1(2k+2;1;t)$$ should instead read $$\int_0^{\infty} dx\,x^{3+4k} e^{-x^2/4t}J_0(x) = \frac{1}{2}(*4t*)^{2k+2}(2k+1)!\,_1F_1(2k+2;1;*-t*)$$ resulting in $$I(t,\epsilon)=*2*t e^{*-t*} \sum_{m\text{ odd}}^\infty (-\epsilon)^{\frac{m-1}{2}} (*4*t)^m m! \, L_m(*t*)$$ which agrees with 6.631.10 of G&R. $\endgroup$ – David T Jul 29 '14 at 22:13
  • $\begingroup$ Also, I looked at doing a similar expansion in powers of $c^{2}$, but I was unable to solve even the term-by-term integrals. Did you have any luck with that expansion? $\endgroup$ – David T Jul 29 '14 at 22:18
  • $\begingroup$ @Sheepiedog: Yeah, you're right: I seem to have misidentified $\alpha$ slightly (should be $1/4t)$. As for the behavior at small $c$, I don't think that approach will actually help: the integral seems to blow up logarithmically (as your reference attests as well) as $c\to0$. So that'll require a different approach. $\endgroup$ – Semiclassical Jul 30 '14 at 3:50
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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

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

enter image description here

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  • $\begingroup$ I'm glad to see progress on the small $c$ case (I'm not the best with log-divergent integrals.) I imagine one can pin down some of those non-divergent terms as well. (The constant term in particular, since it can be absorbed into the log to set the 'scale' of the log.) $\endgroup$ – Semiclassical Jul 30 '14 at 12:26
  • $\begingroup$ @Semiclassical, after wrestling with it for a while I was indeed able to pull out a couple more terms of the asymptotic, including the constant term. This is a tricky integral; the higher-order contributions seem to be mostly nonlocal. Maybe this is common for log-divergent integrals? Dunno. $\endgroup$ – Antonio Vargas Jul 30 '14 at 21:10
  • $\begingroup$ That's a damn fine job. And it sounds about right---log-divergent integrals are something physicists doing field theory have to wrestle with a lot, and have had to develop machinery to deal with it. See for example these notes for an example (where Ei($x$) actually plays a supporting role). $\endgroup$ – Semiclassical Jul 30 '14 at 21:25
  • $\begingroup$ Another example which I'm more personally familiar with is the log-divergence of the elliptic integral $K(m)$ as $m\to 1$. As seen in this problem, the structure of the logarithmic divergence is actually rather nice---but you'd probably never know it if you approximated it term-by-term! $\endgroup$ – Semiclassical Jul 30 '14 at 21:51
  • $\begingroup$ Thanks for the hard work. I'll see if I can work out some expressions for $f(a,b)$ for various cases relating $a$ and $b$. Already I see that for $a>>b$, $f(a,b)\sim\frac{1}{2}\gamma-\frac{1}{2}\ln(a)$. $\endgroup$ – David T Jul 30 '14 at 21:58

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