How to prove the following? $$ \int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13}\cdot\frac{\sin(x/15)}{x/15}\ dx<\pi $$
First of all, here is the Fourier transform I used: $$ \hat{f}(y)=F[f(x)]=\int\limits_{-\infty}^{+\infty}f(x)e^{ixy}dx $$
Well, I tried to solve the problem using convolution. For example, here is the way I used to approach the easier version of the given task $\left(I=\int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\ dx\right)$: $$ \frac{\sin x}{x}=f(x),\ \ \ \frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}=g(x)\\ I=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\hat{f}(y)\hat{g}(y)dy=\frac{1}{2}\int\limits_{-1}^1\hat{g}(y)dy=\frac{1}{2}\int\limits_{-\infty}^{+\infty}\hat{g}(y)dy-\int\limits_{1}^{+\infty}\hat{g}(y)dy=\pi-\dots\\ \left. \begin{aligned} &\hat{g}(y)=\frac{1}{2\pi}\cdot3\pi I_{\left[-\frac{1}{3},\frac{1}{3}\right]}(y)*5\pi I_{\left[-\frac{1}{5},\frac{1}{5}\right]}(y)\\ &\int\limits_{-\infty}^{+\infty}\frac{5}{2}I_{\left[-\frac{1}{5},\frac{1}{5}\right]}(y)dy=1 \end{aligned} \right\}\Rightarrow\int\limits_{-\infty}^{+\infty}\hat{g}(y)dy=3\pi\int\limits_{-\infty}^{+\infty}I_{\left[-\frac{1}{3},\frac{1}{3}\right]}(y)\ dy=2\pi $$ I noticed that the integral $J=\int\limits_{1}^{+\infty}\hat{g}(y)\ dy$ doesn't necessarily always equal zero. However, I haven't really managed to prove that the first time $J>0$ is in the initial case (in the given problem). However, I noticed that: $$\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{13}<1\\ \frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{15}>1$$ I think that is not a coincidence. Could someone help me to end the solution?
The convolution of the indicator functions extends up to the sum of their widths. That is, the support of the convolution $\circledast_k I_{[-a_k,a_k]}$ is $\left[-\sum_ka_k,\sum_ka_k\right]$. The convolution is a non-negative function, so its integral from $1$ to $\infty$ is positive iff its support extends beyond $1$, that is, if $\sum_ka_k\gt1$.