# How to solve $\int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$ or $\int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$

I need to solve either of the following integrals, they are related to each other by integration by parts:

$$I_1 = \int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$$

or

$$I_2 = \int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$$

I have not been able to find them in integral tables and I wonder if it is possible to solve either of these integrals.

Any ideas on how to calculate $I_1$ or $I_2$?

• Are you confident with Bessel and Struve functions? – Jack D'Aurizio Sep 23 '16 at 14:15
• put $x=\sinh(t)$ then you obtain $$I_1=\int_0^{\infty}\exp(-a\sinh(t))\sinh(t)$$ then have a look here dlmf.nist.gov/11.5 and here dlmf.nist.gov/11.4 – tired Sep 23 '16 at 14:17
• No. I know how to solve it in terms of these, but I would like to have a function as a solution, not another integral or infinite sum @JackD'Aurizio – Mencia Sep 23 '16 at 14:17
• @Mencia: $Y_1(z)$ is a function. It also equals an infinite series, just like $e^z$ equals $\sum_{n\geq 0}\frac{z^n}{n!}$. – Jack D'Aurizio Sep 23 '16 at 14:19
• @tired yes, I mentioned they are related by integration by parts. – Mencia Sep 23 '16 at 14:31

## 1 Answer

If you notice it, you will see that this is the form of Laplace transformation. Thus the first integration can be written as the Laplace transformation of (x/root(x^2 +1)) The second integral can be written by integration by parts Which is e^-ax * root(x^2 +1) - integrate(-ae^-at * root(x^2 +1) dx) Which is then equals to e^-ax * root(x^2 +1) + a*laplace of (root(x^2 +1))

Hope this helped

• how does this answer helps the op? – tired Sep 23 '16 at 14:51