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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?



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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
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1 Answer 1

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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.

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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
@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
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