Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

During my research this integral has shown up

$ \frac{1}{2T} \int_{-T}^T \left( 1 - \frac{|\tau|}{T}\right)e^{-\alpha\tau^2}\cos(2\pi f_0 \tau) d\tau$

I tried to solved by taking the real part of a the complex exponential but it didn't work. Any help?

Cheers,

Mikitov

share|improve this question
    
already tried CAS ? –  Luis H Gallardo Jan 6 '11 at 17:39
    
I don't think this integral can be expressed in elementary functions (which is why by the way this doesn't look like homework to me); however it is not too difficult to compute it in terms of the error function (see e.g. en.wikipedia.org/wiki/Error_function). If you would like more details on that, math.stackexchange.com is probably a better place to ask. –  algori Jan 6 '11 at 17:42
    
Yes, no homework in my life anymore... –  mikitov Jan 6 '11 at 21:22
    
What exactly do you want with it? It is just the convolution (up to normalization and linear change of variable) of $e^{-x^2}$ and the Fejer kernel. Of course, there is no algebraic formula (unless you consider integration by parts a.k.a. "expressing in terms of error function" a great step forward) but all reasonable questions shouldn't be hard to answer :). –  fedja Jan 7 '11 at 4:20
add comment

1 Answer

up vote 0 down vote accepted

The limit is 0, the integral from $0$ to $T$ is: $$\frac{i \sqrt{\pi } e^{-\frac{\pi ^2 f^2}{a}} \left(\text{erfi}\left(\frac{\pi f-i a t}{\sqrt{a}}\right)-\text{erfi}\left(\frac{\pi f+i a t}{\sqrt{a}}\right)\right)}{4 a^{1/2}} $$ $$ -\frac{\pi ^{3/2} f e^{-\frac{\pi ^2 f^2}{a}} \text{erfi}\left(\frac{\pi f-i a t}{\sqrt{a}}\right)+\pi ^{3/2} f e^{-\frac{\pi ^2 f^2}{a}} \text{erfi}\left(\frac{\pi f+i a t}{\sqrt{a}}\right)-2 \pi ^{3/2} f e^{-\frac{\pi ^2 f^2}{a}} \text{erfi}\left(\frac{\pi f}{\sqrt{a}}\right)}{4ta^{3/2}} $$ $$ +\frac{\left(1+e^{4 i \pi f t}\right) e^{-t (a t+2 i \pi f)}-2}{4ta}, $$ as per Mathematica.

share|improve this answer
    
Wow, is that hard to read. –  Simon Rose Jan 6 '11 at 21:29
    
$\lim_{T \to \infty}$ ? It does not make any sense to me for any value of alpha. –  mikitov Jan 6 '11 at 21:39
1  
@Simon: I just pasted mathematica's TeXForm. Thank god WZ was there to fix it... @mikitov: The integral converges (for any positive alpha), so when you divide by $T$ it is not at all surprising that the limit is $0.$ –  Igor Rivin Jan 6 '11 at 23:21
    
The thing is, this integral comes from applying the necessary and sufficient conditions for a wide sense stationary process to be ergodic. The independence of alpha makes me feel nervous :-) Thank you in any case. –  mikitov Jan 7 '11 at 13:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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