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I require the following integral involving the modified Bessel functions of the first and second kinds of order one

$$I(a, b, c) = \int_0^{\infty} \frac{\sin(ax)}{x} I_1(bx) K_1(cx) \mathrm{d}x, \quad\text{where} \quad c \ge b$$

For the case $b=c$, Mathematica gives the result

$$I(a, b) = \frac{1}{m} E(m) + \left( 1-\frac{1}{m} \right) K(m), \quad\text{where} \quad m = - \frac{4 b^2}{a^2}$$

where $K(m)$ and $E(m)$ are the complete elliptic integral of the first and second kinds respectively (as defined by Mathematica).

How to solve the general case? Any help is much appreciated.

Please note this is a cross-post from Math.SE where it received no answers after 3 months.

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  • $\begingroup$ The proper venue is math.se ... no answers perhaps means there is no closed form known. $\endgroup$ – Gerald Edgar Dec 8 '15 at 19:18
  • $\begingroup$ what do you want to know: how it depends on $a,b$ and $c$, when they get one becomes large ? Otherwise why would you think there is a 'better' formula than the one you gave for that integral? $\endgroup$ – username Dec 8 '15 at 21:04
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    $\begingroup$ Would a series solution be of use? I have worked out two series, one that converges for a < 1 and one for a > 1, using Mellin transform techniques. I will add them as a solution soon. $\endgroup$ – Tom Dickens Dec 17 '15 at 23:43
  • $\begingroup$ @TomDickens a series solution would certainly be of use. Many thanks in advance. $\endgroup$ – Nigel1 Dec 18 '15 at 9:46
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One can use Mellin transforms to tackle this integral; in this case one obtains series involving hypergeometric functions. Here is a very brief summary of the process. I plan to complete the answer with more details, and with an asymptotic series for large $a$ soon.

Setting $$f(x) = \frac{\sin(ax)}{x}$$ and $$g(x) = I_1(bx) K_1(cx),$$ one has the Mellin transforms $$ F(s) = a^{1-s} \cos \left( \frac{\pi s}{2}\right) \Gamma(s-1) $$ and $$ G(s) = 2^{s-2} \frac{b}{c^{s+1}}\ \Gamma(s/2)\ \Gamma(1+s/2)\ {}_2 F_1\! \left[ \frac{s}{2}, \frac{s+2}{2}; 2 ; \frac{b^2}{c^2} \right].$$

The Parseval theorem for Mellin transforms gives $$ I = \frac{1}{2 \pi i} \int_{k-i\infty}^{k+i\infty} ds\ F(1-s)\ G(s), $$ where $0<k<1.$ The contour can be moved along the real axis to plus or minus infinity, picking up residues from the poles as you go.

Simple poles are found at $s=2k+1, k=0,1,2,\cdots,$ and displacing the contour over these gives a series useful for $a<1$, given by $$ I(a,b,c) = \frac{b}{2} \sum_{k=0}^\infty \frac{a^{2k+1}}{(2k+1)!} \frac{(-1)^k}{c^{2k+2}} 4^k \left(k+\frac{1}{2}\right)\Gamma^2\left( k+\frac{1}{2}\right) {}_2 F_1\! \left[ k+\frac{1}{2}, k+\frac{3}{2} ; 2 ; \frac{b^2}{c^2} \right] .$$

By computing the hypergeometric function for several values of $k$, I find that it can be expressed in each case as a sum of elliptic integrals of the first and second kind, times polynomials, reminiscent of your answer above. (This was done with Mathematica.)

EDIT:

Alternatively, one can displace the contour over the simple poles located at $s = -2 k, k=0,1,2,\cdots,\infty$ to obtain an asymptotic series useful for large $a$, which after some simplification becomes $$ I = \frac{\pi b}{4 c} \left( 1 + \sum_{k=1} (-1)^k \left(\frac{c}{2 a} \right)^{2 k} \frac{ (2 k)!}{k!^2}\ {}_2F_1\!\left[ -k, -k+1;2;\frac{b^2}{c^2} \right] \right). $$ Note that this series, being asymptotic, must be used with care; i. e. if too many terms are used it will diverge.

In this case the hypergeometric series terminate, and can be evaluated as polynomials in $b^2/c^2$. This can lead to relatively simple expressions useful for large $a$.

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