# A troublesome integral $\frac{1}{a(2\pi)^2}\int^{\infty}_{0} \frac{r(b-cr)}{\sqrt{r^2+m^2}}\sin{(ar)}e^{-i\sqrt{r^2+m^2}t-nr} dr$

I've been trying to solve this integral without much luck.

$$\frac{1}{a(2\pi)^2}\int^{\infty}_{0} \frac{r(b-cr)}{\sqrt{r^2+m^2}}\sin{(ar)}e^{-i\sqrt{r^2+m^2}t-nr} dr,$$

or alternatively

$$\frac{1}{a(2\pi)^2}\int^{\infty}_{0} \frac{r(b-cr)}{2i\sqrt{r^2+m^2}}\left(e^{-i(\sqrt{r^2+m^2}t -ar) -nr} - e^{-i(\sqrt{r^2+m^2}t + ar) -nr}\right) dr,$$

where $a,b,c,m,n,t\in\mathbb{R}$ are all positive constants.

I'm really struggling to overcome the $\sqrt{r^2+m^2}$ term in the exponent. I was thinking I could expand this term but then since it is under an integral it doesn't make much sense just choosing a point to expand about. Also don't think the expansion would yield a nice integral anyway. I computed a similar integral earlier where $m=0$ using just a substitution for $r$ and the defninition of the gamma function. However I can't see a way to manipulate this integral into an acceptable form.

I have also been looking in Gradshteyn and Ryzhik for an appropriate integral but can't find one, 3.914 (6) on page 491 (p.540 of the pdf) is the closest I could find but there is no linear term in the exponential.

Can anyone see a method to compute this integral?

• The first thing I would do would be to differentiate inside the integral sign with regard to the parameter t. The second thing I would do would be to let $r=m\sinh x$, and to employ Bessel and Struve functions. Feb 24, 2015 at 0:38
• After that substitution I get an integral of the form, $a\int^{\infty}_{0}\sinh{u}\cosh{u}(c\sinh{u} -b)\left(e^{-imt\cosh{u} - n\sinh{u}} - e^{-imt\cosh{u} - \overline{n}\sinh{u}}\right)$. The constants are not the same as above but are all still in $\mathbb{R}$ except $n\in\mathbb{C}$. So now I have 4 seperate integrals, I have been looking at integral representations of special functions but I cannot find any in this form. I.e a linear combination of $\sinh$ and $\cosh$ in the exponent along with a product of powers of $\sinh$ and $\cosh$. Is their another manipulation required? Feb 24, 2015 at 18:15
• We can lose the $\cosh{u}$ term that is not inside the exponent if we dont differentiate by t prior however I cannot find an integral representation of that form either. Feb 24, 2015 at 18:20
• Notice that some of these four integrals can in their turn be expressed in terms of simpler ones, by differentiation under the integral sign with respect to m and n. Feb 24, 2015 at 19:02

HINT: Substitute $u = it \sqrt{r^2+m^2}, du = \frac{itr}{\sqrt{r^2+m^2}}dr$. Then $r = ((-iu/t)^2-m^2)^{\frac{1}{2}}$ and your integral has the form (Integration Begins at $u=itm$; up to a constant factor):

$\int_{itm}^\infty (b-cr)(e^{(ia-n)r}-e^{(-ia-n)r})e^{-u}du =$

$\int_{itm}^\infty (b-cr) \sum_{k=0}^\infty \frac{1}{k!}((ia-n)^k-(-ia-n)^k)r^ke^{-u}du$.

To compute $r^k$ in dependence on $u$ you can use the generalized binomial Theorem:

$((-iu/t)^2-m^2)^{\frac{k}{2}} = \sum_{l=0}^\infty (\frac{k}{2} over l) (-iu/t)^{k-2l}(-m^2)^l$.

Now you can use the linearity of the sums and the integral and you end up with integrals of the form: $\int_{itm}^\infty u^{q-1} e^{-u} du = I_q$.

This is the incomplete Gamma function, i.e. it holds: $I_q = \Gamma(q,itm)$.

• Inside the first summation is it not meant to be $-(-ia-n)^k$? Feb 24, 2015 at 0:14
• yes, it is meant $-(-ia-n)^k$. Feb 24, 2015 at 10:29