# Definite integral with modified Bessel functions, trigonometric function and a power

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.

• The proper venue is math.se ... no answers perhaps means there is no closed form known. – Gerald Edgar Dec 8 '15 at 19:18
• 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? – username Dec 8 '15 at 21:04
• 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. – Tom Dickens Dec 17 '15 at 23:43
• @TomDickens a series solution would certainly be of use. Many thanks in advance. – Nigel1 Dec 18 '15 at 9:46

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 $a$.