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Given the integral $(1)$

$$\int_{0}^{\infty}e^{-x}\sin^n x\left({1\over x}+{1\over x^2}+{1\over x^3}+\cdots+{1\over x^n}\right)dx =\arctan(n)\tag1$$

An attempt:

Rewrite $(1)$ as

$$\sum_{j=1}^{n}\int_{0}^{\infty}e^{-x}\sin^n x\cdot{\mathrm dx\over x^j}\tag2$$

Maybe if we know the closed form for $(3)$, it could be useful

$$\sum_{j=1}^{n}{1\over x^j}?\tag3$$

Employing $e^{-x}$ series $(2)$ becomes

$$\sum_{j=1}^{n}\int_{0}^{\infty}\sin^n x\left(x^{-j}-x^{1-j}+{x^{2-j}\over 2!}-{x^{3-j}\over 3!}+\cdots \right)\mathrm dx\tag4$$

I guess we may use integration by parts to integrate term by term, it may results in a lengthy process.

How else may integral $(1)$ be prove?

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  • 1
    $\begingroup$ $\sin^nx\left({1\over x}+{1\over x^2}+\cdots+{1\over x^n}\right)=\left(\sin x\over x\right)^n\left(x^n-1\over x-1\right)$, but I'm not sure how much that helps. $\endgroup$ – Barry Cipra Mar 21 '17 at 16:50
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    $\begingroup$ For $n=1$, $n=2$ and $n=3$ the equality seems to be true, but for $n=4$ the integral equals (according to Mathematica) $$\frac{1}{48} \left(\log \left(\frac{625}{17}\right)-8 \arctan(2)+52 \arctan(4)\right)\approx 1.32687$$ which is not equal to $$\arctan(4)\approx 1.32582.$$ $\endgroup$ – mickep Mar 21 '17 at 17:43
  • $\begingroup$ Through the Laplace transform it is straightforward to check that the given identity cannot hold for any even $n\geq 4$. We have a simple pole that contributes with a logarithmic term in the closed form. $\endgroup$ – Jack D'Aurizio Mar 21 '17 at 18:12
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The inverse Laplace transform of $e^{-x}\left(\frac{1}{x}+\ldots+\frac{1}{x^n}\right)$ is given by $$ \mathbb{1}_{s\geq 1}(s)\sum_{k=0}^{n-1}\frac{(s-1)^k}{k!} $$ and the Laplace transform of $\sin^n(x)$ is given by $$ \frac{n!}{s(4+s^2)(16+s^2)\cdots(n^2+s^2)} $$ if $n$ is even and by $$ \frac{n!}{(1+s^2)(9+s^2)\cdots (n^2+s^2)} $$ if $n$ is odd. The given integral boils down to the integral of a rational function over $(1,+\infty)$,
but for $n=4$

$$ \int_{1}^{+\infty}\frac{4!}{s(4+s^2)(16+s^2)}\sum_{k=0}^{3}\frac{(s-1)^k}{k!}\,ds \neq \arctan(4)$$ since the LHS depends on $\log\frac{625}{17}$. Anyway, through the previous representation we may simply compute the given integral through partial fraction decomposition. The given identity holds for $n\in\{1,2,3\}$, but this problem rightfully belongs to the Mathematical coincidence Wikipedia page.

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  • 3
    $\begingroup$ nice counterexample (+1). it is strange that OP didn't check his conjecture for a few more values by a CAS $\endgroup$ – tired Mar 21 '17 at 18:17

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