How to integrate a fraction of sums of exponentials?

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SUMMARY

The integral \(\int_0^\infty \frac{e^{-ax}}{e^{-bx}+e^{-cx}}dx\) can be approached by first clarifying the signs of the constants \(a\), \(b\), and \(c\). If these constants are positive, a substitution such as \(\displaystyle{e^{-ax}} = t\) simplifies the expression. A Taylor expansion is not the only method available; understanding the behavior of the exponential terms is crucial for finding a solution.

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  • Understanding of integral calculus, specifically improper integrals.
  • Familiarity with exponential functions and their properties.
  • Knowledge of substitution techniques in integration.
  • Basic concepts of Taylor series expansions.
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DRJP
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Is it possible to have an solution to this sort of integral? And if not, why not?

[tex]\int_0^\infty \frac{e^{-ax}}{e^{-bx}+e^{-cx}}dx[/tex]

Is a Taylor expansion the only way forward?

Many thanks
David
 
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DRJP said:
Is it possible to have an solution to this sort of integral? And if not, why not?

[itex]\int_0^\infty \frac{e^{-ax}}{e^{-bx}+e^{-cx}}dx[/itex]

Is a Taylor expansion the only way forward?

Many thanks
David

Use [tex ] instead of inline tex if you're not writing a formula on the same line with words.

[tex]\int_0^\infty \frac{e^{-ax}}{e^{-bx}+e^{-cx}}dx[/tex]

looks better and is easier to read.

As for your question, before jumping to series expansions and substitutions, specify if the arbitrary constants are positive or negative. This makes a huge difference on the final result.
Then try to get rid of as many exponentials as possible. You can make the substitution (a,b,c >0) [itex]\displaystyle{e^{-ax}} = t[/itex] and see what you get.
 

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