Encouraged by Abhineet Nayyar

First, some introduction...

• Dirichlet Integral: $\displaystyle\int_0^\infty \dfrac{\sin x}x \,dx= \dfrac{\pi}2$

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• $\displaystyle \int_0^\infty \dfrac{\sin^3x}xdx=\dfrac\pi4$ :

We know $\sin^3x = \dfrac{3\sin x -\sin3x}4$ $\begin{aligned} \int_0^\infty \dfrac{\sin^3x}xdx &=\dfrac14\left[ \int_0^\infty\dfrac{3\sin x}xdx-\int_0^\infty\dfrac{\sin3x}xdx\right]\\&=\dfrac{3\pi}8-\dfrac14\int_0^\infty \dfrac{\sin y}ydy\\&=\dfrac\pi4\end{aligned}$

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• $\displaystyle \int_0^\infty \dfrac{\sin^5x}xdx=\dfrac{3\pi}{16}$

We know $\sin^5x=\dfrac{\sin 5x +20\sin^3x-5\sin x}{16}$ $\begin{aligned}\int_0^\infty\dfrac{\sin^5x}xdx&=\dfrac1{16}\left[\int_0^\infty \dfrac{\sin5x}xdx+20\int_0^\infty\dfrac{\sin^3x}xdx - 5\int_0^\infty\dfrac{\sin x}xdx\right]\\&=\dfrac1{16}\left[\dfrac \pi 2 + 5\pi - \dfrac{5\pi}2\right]\\&=\dfrac{3\pi}{16}\end{aligned}$

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• Since the integrations by parts is from $0\to\infty$ we can ignore the term $uv$ in $\int u\,dv=uv-\int v\,du$

So, we can use the following here(not everywhere). You can use this here because $\lim_{x\to\infty}\int v(x)dx = \lim_{x\to\infty}\dfrac 1{x^n}=0 \text{, for some }n > 0$ and $u(0)=\sin 0=0$ So, $\int_0^\infty u(x)v(x)dx = -\int_0^\infty \left(u'(x)\int v(x)dx\right)dx$

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• $\displaystyle\int_0^\infty \dfrac{sin^3x}{x^3}dx = \dfrac{3\pi}8$

$\displaystyle\begin{aligned}\int_0^\infty \dfrac{\sin^3x}{x^3}dx&=\dfrac32\int_0^\infty\dfrac{\sin^2x\cos x}{x^2}dx\\&=\dfrac32\int_0^\infty\dfrac{2\sin x\cos^2x-sin^3x}xdx\\&=\dfrac32\left[2\int_0^\infty\dfrac{\sin x}xdx-3\int_0^\infty\dfrac{\sin^3x}xdx\right]\\&=\dfrac32\left(\pi-\dfrac{3\pi}4\right)\\&=\dfrac{3\pi}8\end{aligned}$

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Let's do Parts! $\begin{aligned} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx &=\dfrac 54 \int_0^\infty \dfrac{\sin^4x\cos x}{x^4}dx\\&=\dfrac5{12} \int_0^\infty \dfrac{4\sin^3x\cos^2x-\sin^5x}{x^3}dx\end{aligned}$ Using $\cos^2x=1-\sin^2x$ $\begin{aligned} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac5{12}\left[ \int_0^\infty \dfrac{4\sin^3x}{x^3}dx- \int_0^\infty \dfrac{5\sin^5}{x^3}dx\right]\end{aligned}$ Using $\displaystyle\int_0^\infty \dfrac{sin^3x}{x^3}dx = \dfrac{3\pi}8$ , $\begin{aligned} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}8-\dfrac{25}{12} \int_0^\infty \dfrac{\sin^5x}{x^3}dx\end{aligned}$ Repeating the Integration By Parts again, $\begin{aligned} \int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}{8}-\dfrac{125}{24}\left[ \int_0^\infty \dfrac{4\sin^3x}xdx- \int_0^\infty \dfrac{5\sin^5x}xdx\right]\end{aligned}$

Using $\displaystyle \int_0^\infty \dfrac{\sin^3x}xdx=\dfrac\pi4$ and $\displaystyle \int_0^\infty \dfrac{\sin^5x}xdx=\dfrac{3\pi}{16}$

So, $\int_0^\infty \dfrac{4\sin^3x}xdx- \int_0^\infty \dfrac{5\sin^5x}xdx =\pi-\dfrac{15\pi}{16}= \dfrac{\pi}{16}$

$\begin{aligned} \Rightarrow\int_0^\infty \left( \frac{\sin x}x \right)^5 \, dx&=\dfrac{5\pi}8-\dfrac{125\pi}{24\times16}\\&=\dfrac{115\pi}{384}\end{aligned}$

$\displaystyle\Huge \therefore\int_0^\infty \left( \frac{\sin x}x \right)^5 dx =\boxed{\dfrac{115\pi}{384}}$

So, $a+b+1=115+384+1=\boxed{500}$

First, for general indefinite integral

$\displaystyle \int { \dfrac { Sin\left( ax \right) }{ { x }^{ 5 } } dx }$

We use the recursive relations

$\displaystyle \int { \dfrac { Sin\left( ax \right) }{ { x }^{ n } } dx } =-\dfrac { Sin\left( ax \right) }{ \left( n-1 \right) { x }^{ n-1 } } +\dfrac { a }{ \left( n-1 \right) }\displaystyle \int { \dfrac { Cos\left( ax \right) }{ { x }^{ n-1 } } dx }$

$\displaystyle \int { \dfrac { Cos\left( ax \right) }{ { x }^{ n } } dx } =-\dfrac { Cos\left( ax \right) }{ \left( n-1 \right) { x }^{ n-1 } } -\dfrac { a }{ \left( n-1 \right) }\displaystyle \int { \dfrac { Sin\left( ax \right) }{ { x }^{ n-1 } } dx }$

to derive the following relation

$\displaystyle\int { \dfrac { Sin\left( ax \right) }{ { x }^{ 5 } } dx } =\dfrac { 1 }{ 24{ x }^{ 4 } } \left( { a }^{ 3 }{ x }^{ 3 }Cos\left( ax \right) +{ b }^{ 2 }{ x }^{ 2 }Sin\left( ax \right) -2bxCos\left( ax \right) -6Sin\left( ax \right) \right) +\dfrac {{ b }^{ 4 }}{24}\displaystyle\int { \dfrac { Sin\left( ax \right) }{ x } dx }$

Now, using trigonometric identities, we have the following relation

${ \left( Sin\left( x \right) \right) }^{ 5 }=\dfrac { 1 }{ 16 } (10Sin\left( x \right) -5Sin\left( 3x \right) +Sin\left( 5x \right) )$

so that, using both relations, we have

$\displaystyle \int _{ 0 }^{ \infty }{ { \left( \dfrac { Sin\left( x \right) }{ x } \right) }^{ 5 } }dx =Limit\left| f\left( x \right) ,x\rightarrow\infty \right| -Limit\left| f\left( x \right) ,x\rightarrow0 \right| +\dfrac { 115 }{ 192 }\displaystyle \int _{ 0 }^{ \infty }{ \dfrac { Sin\left( x \right) }{ x } } dx$

where $f(x)$ is a sum of terms of the form

$\dfrac { bSin\left( ax \right) }{ c{ x }^{ n } } \quad or\quad \dfrac { bCos\left( ax \right) }{ c{ x }^{ n } }$

where $0<n<5$ , and $a, b, c$ are integers. Using L’Hospital’s rule, all of the terms vanish at either limit, i.e., we have $f(0)=0$ and $f(\infty)=0$ , and so we are left with

$\displaystyle \int _{ 0 }^{ \infty }{ { \left(\dfrac { Sin\left( x \right) }{ x } \right) }^{ 5 } }dx =\dfrac { 115 }{ 192 }\displaystyle \int _{ 0 }^{ \infty }{\dfrac { Sin\left( x \right) }{ x } } dx=\dfrac { 115\pi }{ 384 }$

Note that with this approach, which can be generalized for other $n$ , we really only need to pay attention to the coefficients of the Dirichlet Integral function and ignore the rest, keeping in mind the following detail

$\displaystyle \int _{ 0 }^{ \infty }{ \dfrac { Sin(ax) }{ x } dx } =\displaystyle \int _{ 0 }^{ \infty }{ \dfrac { Sin(x) }{ x } } dx=\dfrac { \pi }{ 2 }$

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Edit : Using the following trigonometric identity, for odd $n$ , for those coefficients

${ \left( Sin(x) \right) }^{ n }=\dfrac { 2 }{ { 2 }^{ n } }\displaystyle \sum _{ k=0 }^{ \frac { n-1 }{ 2 } }{ \left( { \left( -1 \right) }^{ \left( \frac { n-1 }{ 2 } -k \right) }\dfrac { n! }{ k!\left( n-k \right) ! } Sin\left( \left( n-2k \right) x \right) \right) }$

and using the approach outlined above for $n=5$ , this works out to, for odd $n$

$\displaystyle \int _{ 0 }^{ \infty }{ { \left( \dfrac { Sin(x) }{ x } \right) }^{ n } } =\dfrac { \pi }{ { 2 }^{ n }\left( n-1 \right) ! } \displaystyle \sum _{ k=0 }^{ \frac { n-1 }{ 2 } }{ \left( { \left( -1 \right) }^{ k }\dfrac { n! }{ k!\left( n-k \right) ! } { \left( n-2k \right) }^{ n-1 } \right) }$

Yes, this does look like Sharma's result, except for the coefficient on the left side of the $\sum$ . The coefficient here is sufficient, no need to employ a power of $i$ and the $Sine$ function. And, of course , $(n-1)!=\Gamma(n)$

I will try to find an expression for the general integral.

$\displaystyle \int_{0}^{\infty}{{\left(\frac{\sin x}{x}\right)}^n \, dx}$

First let's consider that $n$ is odd. Then,

$\displaystyle \int_{0}^{\infty}{\frac{(e^{ix} - e^{-ix})^n}{(2i)^n x^n} \, dx}$

$\displaystyle - {\left(\frac{i}{2}\right)}^{n-1} \int_{0}^{\infty}{\frac{1}{2i} (1-e^{2ix})^n \frac{dx}{e^{inx} x^n}}$

$\displaystyle - {\left(\frac{i}{2}\right)}^{n-1} \int_{0}^{\infty}{\frac{1}{2i} \sum_{k=0}^{n}{\binom{n}{k}(-1)^k e^{(2k-n)ix}} \frac{dx}{x^n}}$

As the first case is for odd $n$ ,

$\displaystyle - {\left(\frac{i}{2}\right)}^{n-1} \int_{0}^{\infty}{\frac{1}{2i} \sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k (e^{(2k-n)ix} - e^{-(2k -n)ix})} \frac{ dx}{x^n}}$

$\displaystyle - {\left(\frac{i}{2}\right)}^{n-1} \int_{0}^{\infty}{\sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k \sin((n-2k)x)} \frac{dx}{x^n}}$

Substitute $x = \sqrt{u}$

$\displaystyle - \frac{i^{n-1}}{2^n} \int_{0}^{\infty}{\sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k \frac{\sin((n-2k)\sqrt{u})}{\sqrt{u}}} u^{1-n/2 -1} \, du}$

Now, we can use Mellin transform and hence Ramanujan Master Theorem.

Here, for the generating function, we'll use $\displaystyle \frac{\sin(\sqrt{x})}{\sqrt{x}} = \sum_{i=0}^{\infty}{\frac{(-u)^i}{(2i+1)!}}$ . And our generating function would become

$\displaystyle \alpha(u) = \sum_{k=0}^{(n-1)/2}{\binom{n}{k}(-1)^k \sum_{m=0}^{\infty}{\frac{(n-2k)^{2m+1}\Gamma(m+1)}{\Gamma(2m+2)} \frac{(-u)^m}{m!}}}$

$\displaystyle = \sum_{m=0}^{\infty}{\left(\frac{\Gamma(m+1)}{\Gamma(2m+2)} \sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k (n-2k)^{2m+1}}\right) \frac{(-u)^m}{m!}}$

so that $\displaystyle \phi(m) = \frac{\Gamma(m+1)}{\Gamma(2m+2)} \sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k (n-2k)^{2m+1}}$

Now, using RMT, our integral becomes

$\displaystyle = - \frac{i^{n-1}}{2^n} \Gamma\left(\frac{2-n}{2}\right) \phi\left(\frac{n-2}{2}\right)$

$\displaystyle = - \frac{i^{n-1}}{2^n}\Gamma\left(\frac{2-n}{2}\right) \frac{\Gamma\left(\frac{n-2}{2}+1\right)}{\Gamma(n-2+2)} \sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k (n-2k)^{n-2+1}}$

Finally,

$\displaystyle \int_{0}^{\infty}{{\left(\frac{\sin x}{x}\right)}^n \, dx} = - \frac{i^{n-1}}{2^n}\frac{\pi}{\sin\left(\frac{\pi(2-n)}{2}\right) \Gamma(n)} \sum_{k=0}^{(n-1)/2}{\binom{n}{k} (-1)^k (n-2k)^{n-1}}, \text{for odd n}$