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I am trying to show that: $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx = \int_0^\infty \frac{\sin^{2n}x}{x^2}dx$$ for any positive integer $n$. This eqn struck me when I was evaluating $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx $$ and $$ \int_0^\infty \frac{\sin^{2n}x}{x^2}dx$$ for different values of $n$ on Wolfram Alpha, and I noticed that they always came out to be equal.

I verified this eqn to hold from $n=1$ to $n= 26$ on Wolfram Alpha so I generalised it for all positive integer $n$. I tried to prove this using Integration by parts as follows:

\begin{align*} & \int_0^\infty \frac{\sin^{2n}x}{x^2}dx = \int_0^\infty 1.\frac{\sin^{2n}x}{x^2}dx \\ &= \left[ \frac{sin^{2n}x}{x^2}.x\right]_{0}^{\infty}-\int_0^\infty \frac{2n\sin^{2n-1}x.cosx.x^2-2x.sin^{2n}x}{x^4}.xdx\\ &= 0-2n\int_0^\infty \frac{\sin^{2n-1}x.cosx}{x}dx + 2\int_0^\infty \frac{\sin^{2n}x}{x^2}dx \\ &\Rightarrow \int_0^\infty \frac{\sin^{2n}x}{x^2}dx = 2n \int_0^\infty \frac{\sin^{2n-1}x.cosx}{x}dx\\ \end{align*} Now I am stuck at the process of proving: $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx = 2n\int_0^\infty \frac{\sin^{2n-1}x.cosx}{x} dx$$ How do I proceed? Please help.

Arpan Mathur
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1 Answers1

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For the evaluation of $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx\quad\mbox{and}\quad \int_0^\infty \frac{\sin^{2n-1}x\cos x}{x}d x,$$ apply the Power-reduction formulae, $$\sin^{2n-1}x=\frac{1}{4^{n-1}}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{n+k}\sin((2k+1)x),$$ note that $$\sin((2k+1)x)\cos x=\frac{1}{2}\left(\sin((2k+2)x)+\sin(2kx)\right),$$ and use the fact that for $a>0$, $$\int_{0}^{+\infty}\frac{\sin(ax)}{x}\,dx=\frac{\pi}{2}.$$ Hence $$\int_0^\infty \frac{\sin^{2n-1}x}{x}dx= \frac{1}{4^{n-1}}\sum_{k=0}^{n-1}(-1)^{k}\binom{2n-1}{n+k}\int_0^\infty\frac{\sin((2k+1)x)}{x}dx=\frac{n\pi}{4^n(2n-1)}\binom{2n}{n}$$ and $$\int_0^\infty \frac{\sin^{2n-1}x\cos x}{x}dx=\frac{\pi}{2\cdot4^n(2n-1)}\binom{2n}{n}.$$

Arpan Mathur
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Robert Z
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