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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...

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    $\begingroup$ Wolfram gives the value $\frac{5\pi}{16}$ for $n=3$, so the conjecture is false. $\endgroup$ – uniquesolution Mar 27 '17 at 10:30
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    $\begingroup$ ... for $n=4$ we have $\frac{19\pi}{60}$, for $n=5$ we have $\frac{383 \pi }{1152}$. $\endgroup$ – Jack D'Aurizio Mar 27 '17 at 10:48
  • $\begingroup$ In (2) the integral $\int_{-\infty}^{\infty} \dfrac{x^{n-k}\sin^{k}x}{x^2}\, dx$ ought to be $\int_{-\infty}^{\infty} \dfrac{x^{-k}\sin^{k}x}{x^2}\, dx$ which diverges. $\endgroup$ – JanG Mar 27 '17 at 11:22
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    $\begingroup$ All your recent questions are about more or less the same type of integrals (this one being very close to the one you asked 15 hours ago (math.stackexchange.com/q/2204250)). Could you say why you are interested in this type of integrals more than others ? Do you have a textbook with a list of integrals to be computed ? A friendly advice: don't spend too much time on it : till a certain point, you learn analysis techniques, but beyond, it can become rather sterile... $\endgroup$ – Jean Marie Mar 27 '17 at 12:26
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    $\begingroup$ I got the value $\displaystyle - 4\pi\sum_{k=1}^{n}\binom{n}{k}\cdot\sum_{j=1+\lfloor k/2\rfloor}^{k}\binom{k}{j}(-1)^{j}\dfrac{(j-\frac{k}{2})^{k+1}}{(k+1)!}$. $\endgroup$ – JanG Mar 27 '17 at 19:35
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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*}

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