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I know that the improper integral $\int_0^\infty\sin(\sin x)dx$ diverges since 2 ways of summing it give different values. Now my question is whether that divided by $x$ converges.

I have a feeling that it does since it's similar to $\frac{\sin x}x$. I had an idea of using the Abel-Dirichlet criterion since the integral of $\sin(\sin x)$ is bounded (since areas cancel out sort of) and $\frac1x$ is monotonous from $\pi/2$ to $\infty$ but I'm not 100% sure if it works.

Anne Bauval
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mini eden
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1 Answers1

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It works. Define, for $k\geq 1$, $u_k=\int_{k\pi}^{(k+1)\pi} \sin(\sin x)\, \frac{dx}x$. Then $(-1)^k u_k>0$ and $\vert u_{k+1}\vert<\vert u_k\vert $, so that $\sum_k u_k$ is convergent.

PGLR
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