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Let $$I_{\mp n}=\int_{-1}^{1} \frac{x^2}{1\mp e^{\sin(x^n)}} \,dx,~~ n=1,3.$$ We can visualize four integrals: $$I_{+1}, ~~I_{-1}, ~~I_{+3}, ~~I_{-3}.$$ Are these four integrals doable and convergent?

Adrian Keister
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Z Ahmed
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  • Yes and the result write $\frac {\color{red}{42}} k$ where $k$ is a whole number and ${\color{red}{42}}$ is "the Answer to the Ultimate Question of Life, The Universe, and Everything" – Claude Leibovici May 14 '19 at 06:12
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    @ClaudeLeibovici: furthermore, it appears that $k$ is a pentatope number (Wikipedia). –  May 14 '19 at 06:48
  • Thanks Rocherz for the editing. Claude and Yves must be joking! – Z Ahmed May 14 '19 at 07:02
  • All four integrals turn out to be $1/3$ as shown by JG in his solution. It is important to note that $I_{+,1}$ and ${+,3}$ are proper definite integrals. $I{-,1}$ and $I_{-,3}$ are improper definite integrals. $I_{-,1}$ – Z Ahmed May 14 '19 at 12:34
  • All four integrals turn out to be $1/3$ as shown by JG in his solution. It is important to note that $I_{+,1}$ and ${+,3}$ are proper definite integrals. But $I{-,1}$ and $I_{-,3}$ are improper definite integrals. $I_{-,1}$ has integrand which exists only as limit $x \rightarrow 0 $ and it is converges to $1/3$. In the case of $I_{-,3}$, the integrand near $x=0$ goes as $1/x$ so $1/3$ is the principal value of ${I_{-,3}.$ – Z Ahmed May 14 '19 at 12:43

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For odd $n$ as in the problem, $\sin x^n$ is odd, and any odd function $f$ satisfies $$\int_{-1}^1\frac{x^2 dx}{1\mp e^f}=\int_0^1 x^2\left(\frac{1}{1\mp e^{-f}}+\frac{1}{1\mp e^f}\right) dx=\int_0^1 x^2dx=\frac13.$$In the case of $I_{-3}$, the integrand is $\sim-\frac{1}{x}$ for small $x$, so this treatment obtains a Cauchy principal value (specifically, case 1) here with $-a=c=1,\,b=0$).

J.G.
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