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How do you compute $$\int_0^{\infty}\frac{\sin^n x}{x^n}dx$$ for every $n$? Thank you.

For $n=1$ it is widely known and for $n=2$ you might use Plancherel's formula. But I don't know how to do it for $n\geq3.$

Frank Zermelo
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  • the only thing i can think about is to expand it into Taylor series but not sure if you can condense the resulting sum after the integral to anything manageable – gt6989b Mar 08 '18 at 03:22
  • WolframAlpha seems to produce analytic answers for $n \in [8]$ which suggests there is an approach (here they are): $$\left{\frac{1}{2}, \frac{1}{2}, \frac{3}{8}, \frac{1}{3}, \frac{115}{384}, \frac{11}{40}, \frac{5887}{23040}, \frac{151}{630}\right} \pi$$ – gt6989b Mar 08 '18 at 03:28
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    https://math.stackexchange.com/questions/34436/evaluating-the-contour-integral-int-0-infty-frac-sin3xx3-math – saulspatz Mar 08 '18 at 03:30
  • It was answered here, I just found out https://math.stackexchange.com/questions/307510/a-sine-integral-int-0-infty-left-frac-sin-x-x-rightn-mathrmdx – Frank Zermelo Mar 08 '18 at 03:35
  • https://math.stackexchange.com/a/307837/16192 – gt6989b Mar 08 '18 at 03:39

1 Answers1

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I guess, we can try some residual formula.

Let $f(z)=\left(\frac{sin(z)}{z}\right)^{n}$, $n\in \mathbb{N}$, and note that $f(z)$ has n-pole in $z=0$.

For $f(z)=\left(\frac{sin(z)}{z}\right)^{n}=Img\left(\frac{e^{inz}}{z^{n}}\right)$ and $|z|\leq Re^{i\theta}$, $\theta \in [0,\pi]$, we have $$|f(z)| \leq \frac{1}{R^{n}} \to 0,$$ if $R \to \infty$.

So, for $n \in \mathbb{N}$, we set the closed curve $\gamma=\gamma_{R} \wedge \gamma_{2} \wedge \gamma_{\delta} \wedge \gamma_{3},$ where $$\gamma_{R}:|z|=Re^{i\theta}, \quad \theta \in[0,\theta] \quad \textrm{and} \quad R>0;$$ $$\gamma_{2}:z=(t-1)R-t\delta, \quad t\in[0,1] \quad \textrm{and} \quad \delta>0;$$ $$\gamma_{\delta}:|z|=\delta e^{i\theta}, \quad \theta \in [\theta,0];$$ $$\gamma_{3}:z=tR+(1-t)\delta, \quad t\in[0,1];$$

Finally, just note that sin(x) and x are odd, then $\frac{sin(x)}{x}$ is even, futhermore $\left(\frac{sin(x)}{x}\right)^{n}$ is even, so $$\int_{0}^{\infty}{\left(\frac{sin(x)}{x}\right)^{n}\textrm{d}x}=\frac{1}{2}\textrm{Img}\int_{-\infty}^{\infty}{\left(\frac{e^{ix}}{x}\right)^{n}\textrm{d}x}$$.

First, note that f(z) is analytic in $int(\gamma)$, by Cauchy-Goursat theorem, we get $$\int_{\gamma}{f(z)\textrm{d}z}=0.$$ By other side, $$\int_{\gamma}{f(z)\textrm{d}z}=\int_{\gamma_{R}}{f(z)\textrm{d}z}+\int_{\gamma_{2}}{f(z)\textrm{d}z}+\int_{\gamma_{\delta}}{f(z)\textrm{d}z}+\int_{\gamma_{3}}{f(z)\textrm{d}z}$$

Now, let $R \to \infty$ and $\delta \to 0$ and we have by fractional residue formula $$\int_{\gamma_{R}}{f(z)\textrm{d}z} \to 0;$$ $$\int_{\gamma_{R}}{f(z)\textrm{d}z} \to -i\pi(Residue\{f(z)\}|_{z=0});$$ $$\int_{\gamma_{2}}{f(z)\textrm{d}z}+\int_{\gamma_{3}}{f(z)\textrm{d}z} \to \int_{-\infty}^{\infty}{\left(\frac{e^{it}}{t}\right)^{n}\textrm{d}t}$$;

Finally, $$0=-i\pi(Residue\{f(z)\}|_{z=0})+\int_{-\infty}^{\infty}{\left(\frac{e^{it}}{t}\right)^{n}\textrm{d}t}.$$ $$\int_{-\infty}^{\infty}{\left(\frac{e^{it}}{t}\right)^{n}\textrm{d}t}=-\pi(Residue\{f(z)\}|_{z=0}).$$

If, you compute $(Residue\{f(z)\}|_{z=0})$, then $$\int_{0}^{\infty}{\left(\frac{sin(x)}{x}\right)^{n}\textrm{d}x}=\frac{1}{2}\textrm{Img}\int_{-\infty}^{\infty}{\left(\frac{e^{ix}}{x}\right)^{n}\textrm{d}x}=\frac{1}{2}\textrm{Im}(i\pi Residue\{f(z)\}|_{z=0})$$

For example, if $n=1$, $Residue\{f(z)\}|_{z=0})=1$ and we get $$\int_{0}^{\infty}{\left(\frac{sin(x)}{x}\right)^{n}\textrm{d}x}=\frac{\pi}{2}$$

Lucas Galhego
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