Computing Integral of $\frac{\sin^n x}{x^n}$ 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 $n\geq3.$ 

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 
$$\int_{0}^{\infty}{\left(\frac{sin(x)}{x}\right)^{n}\textrm{d}x}=\frac{\pi}{2}$$
