Problem from the 2020 Latvian "Sophomore's Dream" competition

Question

Evaluate $$\int_{-a\pi}^{a\pi} \frac{\cos^5(x)+1}{e^x+1}dx, \quad a \in \mathbb{N}$$

In the beginning, I didn't have any ideas of how to solve this. The one that later came to mind was to try using Feynman's technique, but I couldn't think of the proper function to use for the second variable.

Any ideas?

Not sure, but one directions which looks potentially interesting is to write the cosine as a complex exponent — gt6989b, Mar 02 '20 at 17:47
Another potential approach would be to expand $(1+e^x)^{-1}$ as a geometric series, integrate term-by-term, and sum; the sum that comes out of this might be stickier than the initial integral, though... — Steven Stadnicki, Mar 02 '20 at 17:54
One idea is to write this as a complex contour integral with a $z = e^{ix}$ substitution, but then the $e^x \mapsto z^i$ on the bottom becomes tricky. — Ben Grossmann, Mar 02 '20 at 17:54
What is the direction? What is one asked to do with this expression? — Daniel R. Collins, Mar 03 '20 at 03:04
Strongly related: Evaluating $\int_{-a}^a \frac{f(x)+1}{b^x+1},\mathrm dx$ for a $y$-axis symmetrical function $f$ — Maximilian Janisch, Mar 09 '20 at 22:35

LHF · Accepted Answer · 2020-03-02T18:00:31.953

42

Let $I$ be our integral. Substitute $t=-x \Rightarrow dt = -dx$. Then:

$$I=\int_{-a\pi}^{a\pi} \frac{\cos^5 t+1}{e^{-t}+1}\,dt=\int_{-a\pi}^{a\pi} \frac{e^t(\cos^5 t+1)}{e^{t}+1}\,dt=\int_{-a\pi}^{a\pi} \frac{e^x(\cos^5 x+1)}{e^{x}+1}\,dx$$

Therefore:

$$2I=\int_{-a\pi}^{a\pi} \frac{e^x(\cos^5 x+1)}{e^{x}+1}\,dx+\int_{-a\pi}^{a\pi} \frac{\cos^5 x+1}{e^{x}+1}\,dx=\int_{-a\pi}^{a\pi} (\cos^5 x+1)\,dx$$

$$=\bigg[x + \frac{5}{8} \sin(x) + \frac{5}{48} \sin 3 x + \frac{1}{80} \sin 5 x\bigg]_{-a\pi}^{a\pi}=2a\pi$$

Thus $I=a\pi$.

edited Mar 02 '20 at 18:00

answered Mar 02 '20 at 17:54

LHF

8,491

11

+1 elegant answer Atticus – user577215664 Mar 02 '20 at 18:32
16

More generally, if $f$ is even, then $$\int_{-a}^{a}\frac{f(x)}{e^{x}+1},dx=\frac{1}{2}\int_{-a}^{a}f(x),dx=\int_{0}^{a}f(x),dx.$$ – MathIsFun7225 Mar 02 '20 at 23:23
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@MathIsFun7225 I showed this on this forum about a year ago as one of the first things that I did on here – Maximilian Janisch Mar 09 '20 at 22:28

Problem from the 2020 Latvian "Sophomore's Dream" competition

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