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

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.

• I think using polar coordinates for this one seems like a better idea
– user1173615
Commented Dec 20, 2023 at 8:09
• Perhaps you may try series expansion... Commented Dec 20, 2023 at 8:16

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.

• Yes, I made the problem seem complicated, I will try polar coordinates, thank you for your answer Commented Dec 20, 2023 at 8:20
• How are the bounds changing? The new integral obtained seems $\theta$ independent . Commented Dec 20, 2023 at 9:05
• Is my result below correct? Commented Dec 20, 2023 at 10:05

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

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.

• In fact $\int_{Az}^\infty ue^{-u}du=(1+Az)e^{-Az}$.
– J.G.
Commented Dec 20, 2023 at 10:05
• @J.G. Okay. But my result is correct. Right? Thanks. (,,>﹏<,,) Commented Dec 20, 2023 at 10:10
• Sure, but there's no need to use the incomplete gamma for $s\in\Bbb N$.
– J.G.
Commented Dec 20, 2023 at 10:47
• @J.G. Yes. (,,>_<,,) Commented Dec 20, 2023 at 12:44
• Certainly correct, but one can arrive at the answer by simply using integration by parts. Thanks for completing the answer, though!
– user1173615
Commented Dec 20, 2023 at 14:16