Integrals of combinations of exponential, trigonometric and irrational functions

Question

I want to evaluate the following two (converging) integrals.

1. $$\int_0^\infty \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}\; \cos{kx}\; dx$$
2. $$\int_0^\infty \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}\; x\sin{kx}\; dx$$

Here is my trial. To find these, I expanded the Taylor series of trigonometric functions and then used the Bessel function identity, $$K_\nu(z)=\frac{(z/2)^\nu\,\Gamma(1/2)}{\Gamma(\nu+1/2)}\int_0^\infty e^{-z\cosh{t}}\,\text{sinh}^{2\nu}t\,dt$$

1. Let $x=a \sinh{t}$, then we get $$\int_0^\infty \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}\; \cos{kx}\; dx = \int_0^\infty e^{-a\cosh{t}}\,\cos{(ka\sinh{t})}\,dt\\=\int_0^\infty e^{-a\cosh{t}}\,\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(ka\sinh{t})^{2n}\,dt\\=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(ka)^{2n}\frac{\Gamma(n+1/2)}{(a/2)^n\,\Gamma(1/2)}K_n(a)\\=\sum_{n=0}^\infty \frac{(-1)^n}{n!}(k^2a/2)^{n}K_n(a)$$

2. Similarly, we get $$\int_0^\infty \frac{e^{-\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}\; x\sin{kx}\; dx =ka\sum_{n=0}^\infty \frac{(-1)^n}{n!}(k^2a/2)^{n}K_{n+1}(a)$$

Is there any technique to simplify the above two series? Or, I would be very grateful if you could share some of the good integration skills, ideas, or any advice.

Answer 1

The proposed method can be continued to obtain a simple result. By using an integral representation for the modified Bessel functions: \begin{equation} K_{\nu}\left(z\right)=\tfrac{1}{2}(\tfrac{1}{2}z)^{\nu}\int_{0}^{\infty}\exp\left(-t-\frac{z^{2}}{4t}\right)\frac{\mathrm{d}t}{t^{\nu+1}} \end{equation} The series for the cos-integral reads \begin{align} I_c&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}(k^2a/2)^{n}K_n(a)\\&=\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(\frac{k^2a^2}{4}\right)^n\int_{0}^{\infty}\exp\left(-t-\frac{a^{2}}{4t}\right)\frac{\mathrm{d}t}{t^{n+1}}\\ &=\frac{1}{2}\int_{0}^{\infty}\exp\left(-t-\frac{a^{2}}{4t}-\frac{k^2a^2}{4t}\right)\frac{\mathrm{d}t}{t}\\ &=K_0\left( a\sqrt{1+k^2} \right) \end{align} where the geometric series was summed. This result is ref. (3.961.2) in Gradshteyn and Ryzhick Table.

For the sine integral, the shortest path is probably to differentiate the cosine integral with respect to $k$. The result is \begin{equation} I_s=\frac{ak}{\sqrt{1+k^2}}K_1\left( a\sqrt{1+k^2} \right) \end{equation} Alternatively, a direct calculation of these integrals is obtained by adapting the answer of this question to the present case.