- #1

divB

- 87

- 0

I just tried to solve the following integral:

[tex]

\int_{-\infty}^{\infty} \frac{2}{1+(2\pi t)^2} \mathrm{sinc}(2t) dt

[/tex]

My approach is: Convert both to Fourier domain and the multiplication becomes a convolution. Because of Parsevals theorem, I can either integrate in time or frequency domain. Because of linearity, I can then put the outer integral inside:

[tex]

\int_{f=-\infty}^{\infty} \int_{\nu=-\infty}^{\infty} e^{-|\nu|} \frac{1}{2} \Pi\left(\frac{f-\nu}{2}\right) d\nu df \\

= \int_{\nu=-\infty}^{\infty} e^{-|\nu|} \frac{1}{2} \underbrace{\int_{f=-\infty}^{\infty}\Pi\left(\frac{f-\nu}{2}\right) df}_{2} d\nu

[/tex]

It can be clearly seen, that only the double-sided exponential stays whose integral is 2.

So my total solution is 2.

However, Wolfram alpha gives me (e-1)/e:

http://www.wolframalpha.com/input/?i=integral+2%2F%281%2B%282*pi*t%29^2%29+*+sin%28pi*2*t%29%2F%28pi*2*t%29%2Ct%2C-infinity%2Cinfinity

Can anyone explain where I did wrong?

Thanks,

div

PS: The sinc is defined as normalized, i.e. [itex]\mathrm{sinc}(t)=\sin(\pi t)/\pi t[/itex] ...