I was working on my project, and I came across this integration. After googling, I realized that this is an Exponential Integral. And I'm pretty much new to this. Here is the integration

\begin{equation} \int_{R_{min}}^{R_{max}} \frac{e^{-\alpha r}}{r}dr \end{equation}

Here $R_{max}$ & $R_{min}$ is a real numbers like 160 and 50 respectively.

$\alpha$ is a complex number $\alpha = i \frac{2 \pi}{\lambda}$ , here $i$ is $\sqrt{-1}$ and $\lambda$ is a real number .

I have no idea how to evaluate this integral. In my project, I've expressed all the results in terms of $\alpha$. Can you give me any result that can be represented as $\alpha$. Like \begin{equation} \int_{R_{min}}^{R_{max}} e^{-\alpha r}dr \\ =\dfrac{\mathrm{e}^{-\alpha R_{min}}}{\alpha}-\dfrac{\mathrm{e}^{-\alpha R_{max}}}{\alpha} \end{equation}

This above integration is represented in terms of $\alpha$, and I can simply write the code in MATLAB.

Can you give me some results like this one?

  • $\begingroup$ Is $\lambda$ a wavelength? Is this a wave propagation or scattering problem? If so, are you sure this isn't meant to be a three dimensional integral? $\endgroup$ Jun 14, 2020 at 19:22
  • $\begingroup$ @JohnBarber Yes, it is a wavelength. But this is associated with a drone location and represented in spherical coordinate and the three coordinates are independent that's why I've separated them and I have a single integral $\endgroup$ Jun 15, 2020 at 5:39

1 Answer 1


By definition of the exponential integral: $$ \int_{x_1}^{x_2}\frac{e^{-at}}tdt=E_1(a x_1)-E_1(a x_2). $$

  • $\begingroup$ Thanks for your answer, but I'm curious to know how did you calculate this one from the definition? I mean to say in the definition the complex argument is in the limit then how to come up with this solution? $\endgroup$ Jun 14, 2020 at 19:33
  • $\begingroup$ Use substitution $u=at$ in $\int_x^\infty\frac{e^{-at}}tdt$. $\endgroup$
    – user
    Jun 14, 2020 at 19:49
  • $\begingroup$ Just curious to know how to do the same integration when I have $\begin{equation} \int_{R_{min}}^{R_{max}} \frac{e^{\alpha r}}{r}dr \end{equation}$. Just negative value is now positive . $\endgroup$ Jun 15, 2020 at 8:13
  • $\begingroup$ @AmartyaRoy This results in the replacement $a\to -a$ in both sides of the equation. $\endgroup$
    – user
    Jun 15, 2020 at 9:21
  • 1
    $\begingroup$ One should of course be careful. The definition of the exponential integral assumes that $\Re(a)>0$. You should just do not forget that $E_1(z)$ is the analytic continuation of the integral on the whole complex plane. Therefore the given formula is applicable for arbitrary values of $a$. There can be however issues with numerical evaluation of $E_1(z)$. I cannot help in this issue but it is certainly solved. $\endgroup$
    – user
    Jun 15, 2020 at 10:23

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