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I friend of mine gave me the following integral to calculate:

[tex]\int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}[/tex]

This can de done by first doing partial fractions which gives you

[tex]\int_0^\infty \frac{1}{1+e^{bx}}-\frac{1}{1+e^{ax}}[/tex]

then we can calculate each one by substituting [tex]t=1+e^{ax}[/tex] (and similarly for the other term). Hence

[tex]dx=\frac{dt}{a(t-1)}[/tex],

and finally after doing partial fraction again after the substitution, it boils down to

[tex]

\frac{(a-b)\ln 2}{ab}+\lim_{x\rightarrow\infty}\ln\frac{(1+e^{ax})^b}{(1+e^{bx})^a}

=\frac{(a-b)\ln 2}{ab}[/tex]

where the second term disappears. The calculation is straightforward, though quite messy. This had been on an actuarial exam a few years ago where they were actually expecting you to calculate it in under 2 minutes.

I've been trying to figure out if there is an "easier" way to see the answer, but haven't come up with anything useful. I can't convert it into any known integral from Fourier analysis or see how to apply the calculus of residues or something similar. Any ideas?

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# Easier way to do an integral

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