# Simplification of the exponential integral to a closed form

Can the following integral be simplified to a closed form expression?

$$f = e^{yE}\int_{-z}^{z} \frac{e^{\Gamma \beta}}{\beta A + E} \,d\beta - e^{-yE}\int_{-z}^{z} \frac{e^{\xi \beta}}{\beta A + E}\, d\beta$$

I have tried solving the problem and I am getting solution in term of the exponential integral: $$\int e^t/t\,dt = \operatorname{Ei}(t)$$. Further, I tried approximating the exponential integral using the Taylor series expansion of $$e^t/t$$. However, the solution is quite cumbersome.

Is there a solution to the original integration or an approximation without using the $$\operatorname{Ei}(t)$$?

• Are you supposed to know about exponential integrals ? – Claude Leibovici Feb 14 at 8:07
• @ClaudeLeibovici Thank you for the solution you have provided below. I was looking for a closed form solution without the exponential integrals. Or any approximation of closed of exponential integrals. – Chandan Pradhan Feb 14 at 23:37
• Have a look to my edit. – Claude Leibovici Feb 15 at 4:11

Consider $$I=\int \frac {e^{ax}} {bx+c}\,dx$$ Let $$bx+c=t\implies x=\frac{t-c}{b}\implies dx=\frac{dt}{b}$$ which makes $$I=\frac{e^{-\frac{a c}{b}}}{b}\int \frac{e^{\frac{a t}{b}}}{t}\,dt$$ Make now $$\frac{a t}{b}=y$$ to end with something like $$\int \frac {e^y} y \,dy=\text{Ei}(y)$$ Back to $$x$$, $$I=\frac{e^{-\frac{a c}{b}} }{b}\,\text{Ei}\left(\frac{a c}{b}+a x\right)$$
In terms of approximation, for large values of $$y$$, you could use $$\text{Ei}(y)=\frac{e^y}y\sum_{n=0}^\infty \frac {n!}{y^n}$$ and truncate it.
Similarly, for small values of $$y$$, $$\text{Ei}(y)=\gamma+\log(y)+\sum_{n=1}^\infty \frac {y^n} {n \,n!}$$ which, I suppose, is what you used