POTW for University Students week 13

[veronica1999]

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?

 

[Sudharaka]

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]

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 $$\int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}\,dx = \frac{-2\pi ie^{\pi ai}}{1-e^{2\pi ai}} = \frac\pi{{\scriptscriptstyle\frac1{2i}}(e^{\pi ai} - e^{-\pi ai})} = \frac\pi{\sin \pi a}\,.$$