Numerical integration of $E_1(x)$

Question

I want to solve the following integral for $\gamma_0$:

$$\int_{\gamma_0}^\infty \frac{1}{t}e^{-at} dt = c$$

for the specific values $a = 0.01$ and $c = 12.1$. As I understand, this is a variant of the exponential integral function: $E_1(x) = -E_i(-x)$. Wolfram doesn't seem capable of solving this kind of thing. What are my options?

Answer 1

The change of variable $a\,t=s$ changes the integral into $$\int_{a\gamma_0}^\infty \frac{1}{s}e^{-s}\,ds = \Gamma(0,a\,\gamma_0),$$ where $\Gamma(z,s)$ is the incomplete $\gamma$ function. To solve the equation $\Gamma(0,z)=12.1$ you can use Wolfram's function FindRoot. $z$ is of the order of $10^{-6}$, which makes computations a little bit shaky. The following code gives the solution in Mathematica. Try it in Wolframs's cloud.

In[38]:= FindRoot[Gamma[0, z] == 121/10, {z, 1}, 
 WorkingPrecision -> 40, AccuracyGoal -> Infinity, PrecisionGoal -> 30]

Out[38]= {z -> 
  3.121451176974789668567944143395210843560*10^-6 - 
   7.212405063707991989209341377111245941663*10^-51 I}

The options are there to give a sufficiently accurate answer. Observe that the answer includes a tiny but not $0$ imaginary part, which is an artifact of the method used by Mathematica.