Evaluate $\int\limits_{-\infty}^{\infty}\frac{\sin x}{x}\cdot\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}dx$ I would greatly appreciate it if someone could find a mistake in my solution to the following problem.

Evaluate the integral
  $$
I=\int\limits_{-\infty}^{+\infty}\frac{\sin x}{x}\cdot\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}dx
$$

I know that it must equal $\pi$. However, I can't understand what is wrong in my calculations. I used the following Fourier transform:
$$
\hat{f}(y)=F[f(x)]=\int\limits_{-\infty}^{+\infty} f(x)e^{ixy}dx
$$
Thus, we have ($I$ is an Indicator function):
$$
F\left[\frac{\sin \frac{x}{a}}{\frac{x}{a}}\right]=a\pi\cdot I_{\left[-\frac{1}{a}, \frac{1}{a}\right]}(y)
$$
Two other formulas that I used (The sign '$*$' is convolution):
$$
\begin{aligned}
&\int\limits_{-\infty}^{+\infty} f(x)g(x)dx=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} \hat{f}(y)\hat{g}(y)dy\\
&F[fg]=\frac{1}{2\pi}F[f]*F[g]
\end{aligned}
$$
Also, I used one of the properties of convolution:
$$
\text{if}\ \ \ \int\limits_{-\infty}^{+\infty} g(x)dx=1\ \ \ \text{then}\ \ \ \int\limits_{-\infty}^{+\infty} f(x)*g(x)dx=\int\limits_{-\infty}^{+\infty} f(x)dx
$$

So, here is my solution:
  $$
\begin{aligned}
&\text{let}\ \ \ \frac{\sin x}{x}=f(x)\ \ \ \text{and}\ \ \ \frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}=g(x)\\
&I=\int\limits_{-\infty}^{+\infty}f(x)g(x)dx=\int\limits_{-\infty}^{+\infty} \hat{f}(y)\cdot\frac{1}{2\pi}\cdot\hat{g}(y)dy=\frac{1}{2}\int\limits_{-\infty}^{+\infty} I_{[-1, 1]}(y)\cdot\hat{g}(y)dy\\
&\hat{g}(y)dy=F\left[\frac{\sin\frac{x}{3}}{\frac{x}{3}}\cdot\frac{\sin\frac{x}{5}}{\frac{x}{5}}\right]=\frac{1}{2\pi}\cdot 3\pi I_{\left[-\frac{1}{3}, \frac{1}{3}\right]}(y)*5\pi I_{\left[-\frac{1}{5}, \frac{1}{5}\right]}(y)=\\
&=3\pi I_{\left[-\frac{1}{3}, \frac{1}{3}\right]}(y)*\frac{5}{2}\cdot I_{\left[-\frac{1}{5}, \frac{1}{5}\right]}(y)\Rightarrow\text{Here I applied that convolution property}\Rightarrow\\
&\Rightarrow \int\limits_{-\infty}^{+\infty}\hat{g}(y)dy=\int\limits_{-\infty}^{+\infty} 3\pi\cdot I_{\left[-\frac{1}{3},\frac{1}{3}\right]}(y)dy=3\pi\cdot\frac{2}{3}=2\pi
\end{aligned}
$$
  Therefore, we finally get
  $$
I=\frac{1}{2}\int\limits_{-\infty}^{+\infty} I_{[-1,1]}(y)\cdot 2\pi dy=2\pi 
$$

So, what's wrong?

What’s wrong is that you replaced $\hat g(y)$ in the integrand by $\int_{-\infty}^\infty\hat g(y)\mathrm dy$, and there’s no reason why you should be able to do that. What you can do, though, is to argue that the convolution of two rectangular pulses of widths $\frac13$ and $\frac15$ has width less than $1$, so you can omit the indicator function for $[-1,1]$ and write
$$
I=\frac12\int\limits_{-\infty}^{+\infty} I_{[-1, 1]}(y)\cdot\hat g(y)\mathrm dy=\frac12\int\limits_{-\infty}^{+\infty} \hat g(y)\mathrm dy=\frac12\cdot2\pi=\pi\;.
$$
