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POTW for University Students week 13

veronica1999

Member
Jun 4, 2012
63
I know this problem is way beyond my level, but I worked really hard. (1 full week)
After seeing the solution I see my work is nonsense.
Is it all rubbish or is there a tiny bit that can be saved?
 

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Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I know this problem is way beyond my level, but I worked really hard. (1 full week)
After seeing the solution I see my work is nonsense.
Is it all rubbish or is there a tiny bit that can be saved?
Hi veronica1999, :)

When you substitute, \(t=e^{ax}\) the limits of integration should be changed appropriately.

As \(x\rightarrow\infty\), \(t\rightarrow\infty\) and as \(x\rightarrow -\infty\), \(t\rightarrow 0\). Also, \(dt=ae^{ax}\,dx=at\,dx\). Therefore after substitution the integral becomes,

\[\int_{-\infty}^{\infty}\frac{e^{ax}}{1+e^x}\,dx=\frac{1}{a} \int_{0}^{ \infty}\frac{1}{1+\sqrt[a]{t}}dt\]

I cannot guarantee that this approach would work. However this is one of the mistakes that you have done at the very beginning.

Kind Regards,
Sudharaka.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
To my mind, the best way to evaluate $\displaystyle \int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}\,dx$ is to use contour integration. Admittedly, this means that you need to know the residue theorem, but on the other hand it avoids the use of beta and gamma functions (which I dislike).

The idea is to evaluate $\displaystyle \oint_C \frac{e^{az}}{1+e^z}\,dz$, where $C$ is the contour that goes anticlockwise round the rectangle with vertices at $\pm R$ on the real axis and the points $\pm R + 2\pi i.$ When $R$ is large, the integrals along the two vertical sides of the rectangle become small (because of the condition $0<a<1$). The sum of the integrals along the two horizontal sides is $\displaystyle \int_{-R}^R \frac{e^{ax}(1-e^{2\pi ai})}{1+e^x}\,dx.$ There is just one singularity of the integrand inside the contour, at the point $\pi i$, and the residue there is $-e^{\pi ai}.$

Letting $R\to\infty$ and using the residue theorem, you find that $$\int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}\,dx = \frac{-2\pi ie^{\pi ai}}{1-e^{2\pi ai}} = \frac\pi{{\scriptscriptstyle\frac1{2i}}(e^{\pi ai} - e^{-\pi ai})} = \frac\pi{\sin \pi a}\,.$$
 
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