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

$$ I=\frac12\int\limits_{-\infty}^{+\infty} I_{[-1, 1]}(y)\cdot\hat g(y)\mathrm dy=\frac12\int\limits_{-\infty}^{+\infty} \hat g(y)\mathrm dy=\frac12\cdot2\pi=\pi\;. $$

| cite | improve this answer | |
  • 1
    $\begingroup$ God, how haven't I noticed that?! Thank you very much! $\endgroup$ – Bonrey Feb 29 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.