Integrals of the form $e^{-A\sqrt{x^2+B}}$

Question

I now have an integral which can be abbreviated to the form：

$$ e^{-Ar} $$

where $r=\sqrt{x^2+y^2+z^2}$. I need to perform the xy area integral of this function

$$ \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy e^{-A\sqrt{x^2+y^2+z^2}} $$

I found a lot of integrals in the form $e^{-A\sqrt{x}}$ on the Internet, but there is no integral in the form $e^{-A\sqrt{x^2+B}}$.

I hope to know how I should operate on this type of integral, since the actual function is more complicated than this one. Thanks in advance.

Answer 1

The integral in Cartesian coordinates looks fairly complicated, to say the least. If we change to polar coordinates, the integral becomes

$$ I = \int_0^{2\pi} \int_0^{\infty} r e^{-A\sqrt{r^2+z^2}} dr d\theta$$

Now, substitute $ r^2+z^2 = v$ and proceed. It will become an integral of the form $e^{-A\sqrt v}$, which you have said you know how to evaluate. Note that $z$ behaves like a constant.

Answer 2

I want to "complete" the accepted answer of Scipio. $$\begin{align} &\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy e^{-A\sqrt{x^2+y^2+z^2}}\\ &\text{(Change to polar coordinates.)}\\ &= \int_0^{2\pi} \int_0^{\infty} r e^{-A\sqrt{r^2+z^2}} dr d\theta\\ &\text{(Let $v=r^2+z^2.$)}\\ &=\frac12\int_0^{2\pi} \int_{z^2}^{\infty} e^{-A\sqrt{v}} dv d\theta\\ &=\pi\int_{z^2}^{\infty} e^{-A\sqrt{v}} dv\\ &(\text{Let }u=A\sqrt v)\\ &=\frac{2\pi}{A^2}\int_{Az}^\infty ue^{-u}du\\ &=\frac{2\pi}{A^2}\Gamma(2,Az) \end{align}$$ where $\Gamma(s,x)$ is the incomplete gamma function which is a well-known special function.

Answer 3

If you want to compute $$I=\int e^{-A\sqrt{x^2+B}}\,dx$$ but do not plan to integrate it with a lower bound equal to $0$, you could expand as a series.

Assuming $x>0$, you will have $$e^{-A\sqrt{x^2+B}}=e^{-Ax}\Bigg(1+\sum_{n=0}^\infty \frac {P_n(x)}{x^{2n+1}}\, B^n\Bigg)$$ with $$P_n(x)=\frac {-2 (n-1) (2 n-3)\, P_{n-1}+A^2\,P_{n-2} } {4 n(n-1) }$$ with $P_0(x)=1$ and $P_1(x)=-\frac{A}{2 x}$.

This would lead to a series of exponential integral functions.

For example, for the term in $B^2$ $$\int \frac{A e^{-A x} (A x+1)}{8 x^3}\,dx= -\frac{A}8 \Bigg(\frac{A x+1}{2 x^2}\,e^{-A x} +\frac{A^2}{2}\,\, \text{Ei}(-A x) \Bigg)$$