Hi, this is not really a homework problem, but something that came up during my research, upon trying to integrate some empirical function. This function consists of many terms, but specifically, there is a term containing double exponential functions which is giving me some trouble.(adsbygoogle = window.adsbygoogle || []).push({});

If anyone could please assist in helping me with this problem my appreciation would approach infinity :)

1. The problem statement, all variables and given/known data

I want to compute the integral of an empirical relation with respect to x, which I will refer to as y. Here here a, b and c are constant and x>0.

2. Relevant equations

\begin{equation} y = \int_{x_1}^{x_2} \frac{e^{-ax}}{e^{b+e^{-cx}}} dx \end{equation}

3. The attempt at a solution

By using the properties of exponential functions, I simplify the relation:

\begin{equation} y = \int_{x_1}^{x_2} e^{-ax}e^{-b} e^{-e^{-cx}} dx \end{equation}

I have become aware that if a = c, i could make the substitution:

\begin{equation} u = e^{-ax} \end{equation}

and it would be possible to obtain an analytical solution, however this is not the case and I have become stuck in trying to solve this problem. I have also used Mathematica & Matlab to try and obtain symbolic solutions, however both these packages fail (I realize this is hardly an substitution to using analysis, but I was examining different avenues).

Is it perhaps that an analytical solution does not exists? I have plotted the function and it is continuous and reasonably I expected an analytical integral to be possible. Please assist.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Integrating double exponential terms in a function

**Physics Forums | Science Articles, Homework Help, Discussion**