5
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ God, how haven't I noticed that?! Thank you very much! $\endgroup$
    – Bonrey
    Commented Feb 29, 2020 at 17:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .