Another Improper Integral Problem

Evaluate $\displaystyle\int_1^\infty\frac{dx}{e^{x+1}+e^{3-x}}$. (Source: Putnam)

Admittedly, this problem took me an embarrasingly long time to solve (3 hours). Nonetheless, I had fun playing with this particular integral.

For this particular problem, it actually turns out that a symmetric nature is extremely helpful in successfully integrating. To achieve this, we will let $u=x-1$.

\[\int_1^\infty\frac{dx}{e^{x+1}+e^{3-x}}=\int_0^\infty\frac{du}{e^{u+2}+e^{2-u}}.\]

We can then factor $e^2$ out of the denominator:

\[\int_0^\infty\frac{du}{e^{u+2}+e^{2-u}}=\frac{1}{e^2}\int_0^\infty\frac{du}{e^u+e^{-u}}.\]

To get rid of the $e^{-u}$, we multiply the integrand by $\dfrac{e^u}{e^u}$:

\[\frac{1}{e^2}\int_0^\infty\frac{du}{e^u+e^{-u}}=\frac{1}{e^2}\int_0^\infty\frac{e^u}{\left(e^u\right)^2+1}\,du.\]

Now we can make the the substitution $v=e^u$!

\[\begin{aligned} \frac{1}{e^2}\int_0^\infty\frac{e^u}{\left(e^u\right)^2+1}\,du&=\frac{1}{e^2}\int_1^\infty\frac{dv}{v^2+1}\\ &=\frac{1}{e^2}\left[\arctan v\right]^\infty_1\\ &=\frac{1}{e^2}(\arctan\infty-\arctan1). \end{aligned}\]

We know $\displaystyle\lim_{x\to\infty}\arctan{x}=\frac{\pi}{2}$; therefore, our answer is

\[\begin{aligned} \frac{1}{e^2}(\arctan\infty-\arctan1)&=\frac{1}{e^2}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\\ &=\boxed{\frac{\pi}{4e^2}}. \end{aligned}\]

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