$\int_0^{\infty}\frac{\sin^nx}{x^n}dx$, $n>1$, by integration by parts.

Asked Oct 01 '22 at 13:40

Active Oct 01 '22 at 16:58

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My work with integration by parts, IBP:

For $n=2$, $1$ times IBP: $$\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx=\int_0^{\infty}\frac{\sin2x}{x}\mathrm dx=\frac{\pi}{2}.$$

For $n=3$, $2$ times IBP: $$\int_0^{\infty}\frac{\sin^3x}{x^3}\mathrm dx=\int_0^{\infty}\frac{\frac{3}{8}(3\sin3x-\sin x)}{x}\mathrm dx=\frac{3\pi}{8}.$$

For $n=4$, $3$ times IBP: $$\int_0^{\infty}\frac{\sin^4x}{x^4}\mathrm dx=\int_0^{\infty}\frac{\frac{2}{3}(2\sin4x-\sin2x)}{x}\mathrm dx=\frac{\pi}{3}.$$

Question: Can we generalize this? After $(n-1)$ times IBP, can we have $$\int_0^{\infty}\frac{\sin^nx}{x^n}\mathrm dx=\int_0^{\infty}\frac{p_n(x)}{x}\mathrm dx=K_n\pi?$$ Then, what is $p_n(x)$? What is $K_n$?

edited Oct 01 '22 at 16:58

Mittens

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asked Oct 01 '22 at 13:40

Bob Dobbs

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What is IBP, perchance? – Chris Leary Oct 01 '22 at 14:06
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@ChrisLeary Integration By Parts – Sharky Kesa Oct 01 '22 at 14:47
@SharkyKesa Thanks. $K_n$ is shown in that link. It is very complicated. – Bob Dobbs Oct 01 '22 at 14:52
Perhaps better would be integration by parts to find reduction formulas for $\int\frac{\sin^n x}{x^m},dx$ and $\int\frac{\cos^n x}{x^m},dx$. – GEdgar Oct 01 '22 at 17:05
And bye induction, we go to infinty. – Bob Dobbs Oct 01 '22 at 18:47

$\int_0^{\infty}\frac{\sin^nx}{x^n}dx$, $n>1$, by integration by parts.

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