Correct $\int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\left(1-{\sin x\over x}\right)^n={\pi\over n+1}?$ Consider the integral $(1)$

$$\text{Conjecture}\ \int_{-\infty}^{+\infty}{\mathrm dx\over x^2}\left(1-{\sin x\over x}\right)^n=\color{blue}{\pi\over n+1}\tag1$$
  Where $n\ge1$

Required help to prove conjecture $(1)$
An attempt:
Apply binomial series to $(1)$, then we have 
$$\sum_{k=0}^{n}(-1)^k{n\choose k}\int_{-\infty}^{+\infty}{x^{n-k}\sin^k x\over x^2}\mathrm dx\tag2$$
We can apply the trick of $(3)$
$$\int_{-\infty}^{+\infty}f(x){\sin^2 x\over x^2}\mathrm dx=\int_{0}^{\pi}f(x)\mathrm dx\tag3$$
then $(2)$ becomes
$$I_k=\sum_{k=0}^{n}(-1)^k{n\choose k}\int_{0}^{\pi}{x^{n-k}\sin^{k-2} x}\mathrm dx\tag4$$
I guess we could apply integration by parts to $(4)$ but seem difficult...
 A: If we define the Fourier transform via
\begin{equation*}
\hat{f}(\xi) = \int_{-\infty}^{\infty}e^{-i\xi x}f(x)\, dx
\end{equation*}
we get 
\begin{equation*}
\int_{-\infty}^{\infty}\dfrac{1}{x^2}\left(1-\dfrac{\sin x}{x}\right)^n \, dx = \hat{f}(0)
\end{equation*}
where
\begin{equation*}
f(x) = \dfrac{1}{x^2}\left(1-\dfrac{\sin x}{x}\right)^n.
\end{equation*}
The rest of the answer is concentrated on determining $\hat{f}(0).$
Write
\begin{equation*}
x^2f(x) = 1+\sum_{k=1}^{n}\binom{n}{k}(-1)^{k}g_{k}(x)\tag{1}
\end{equation*}
where
\begin{equation*}
g_{k}(x) = \dfrac{\sin^kx}{x^k}.
\end{equation*}
It is well known that $\hat{g_{1}}(\xi) =\pi(H(\xi +1)-H(\xi -1))$, where $H$ is the Heaviside function.
For $k\ge 2$ we study
\begin{equation*}
x^kg_{k}(x) = \sin^kx = (2i)^{-k}\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}e^{i(k-2j)x}.
\end{equation*}
Hence
\begin{equation*}
i^{k}\hat{g_{k}}^{(k)}(\xi) = 2\pi(2i)^{-k}\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}\delta(\xi-k+2j).\tag{2}
\end{equation*}
Then we integrate (2) $k$ times. We have to add a polynomial $p_{k}(\xi)$.
\begin{equation*}
\hat{g_{k}}(\xi) = p_{k}(\xi)+2\pi\cdot(-2)^{-k}\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}\dfrac{(\xi-k+2j)^{k-1}}{(k-1)!}H(\xi-k+2j).\tag{3}
\end{equation*}
But for $k\ge 2$ $g_{k}\in L_{1}({\mathbb{R}})$. According to Riemann-Lebesgue's lemma $\displaystyle\lim_{\xi\to -\infty}\hat{g_{k}}(\xi) = 0$. But for $\xi < 0$ all Heaviside terms in (3) are $0$.
Consequently $\displaystyle \lim_{\xi\to -\infty}p_{k}(\xi) = 0$
which means that  $p_{k}(\xi) = 0$ for all $\xi$. We notice that even $\hat{g_{1}}(\xi)$ is included in (3).
We are now prepared to return to (1). Fourier transformation yields
\begin{gather*}
-\hat{f}^{''}(\xi) = 2\pi\delta(\xi) + \sum_{k=1}^{n}\binom{n}{k}(-1)^{k}\hat{g_{k}}(\xi)=\\[2ex]
2\pi\delta(\xi) + 2\pi\sum_{k=1}^{n}\binom{n}{k}(-1)^{k}(-2)^{-k}\cdot\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}\dfrac{(\xi-k+2j)^{k-1}}{(k-1)!}H(\xi-k+2j).
\end{gather*}
In the next step we integrate twice, change the signs and remove a polynomial of degree one. We get
\begin{equation*}
\hat{f}(\xi) = -2\pi\xi H(\xi)- 2\pi\sum_{k=1}^{n}\binom{n}{k}2^{-k}\cdot\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}\dfrac{(\xi-k+2j)^{k+1}}{(k+1)!}H(\xi-k+2j).
\end{equation*}
Finally we put $\xi = 0$.
\begin{gather*}
\hat{f}(0) = - 2\pi\sum_{k=1}^{n}\binom{n}{k}2^{-k}\cdot\sum_{j=0}^{k}\binom{k}{j}(-1)^{j}\dfrac{(2j-k)^{k+1}}{(k+1)!}H(2j-k)=\\[2ex]
- 4\pi\sum_{k=1}^{n}\binom{n}{k}\cdot\sum_{j=1+\lfloor k/2\rfloor}^{k}\binom{k}{j}(-1)^{j}\dfrac{(j-\frac{k}{2})^{k+1}}{(k+1)!}.
\end{gather*}
