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I tried evaluating the integral but maybe there's an easier way. Please help.

Here is what I did:

$\begin{aligned}\lim_{t\to\infty}\frac1t\int_0^t \sin(\alpha x)\cos(\beta x)dx&=\lim_{t\to\infty}\frac1t\int_0^t\frac12(\sin(\alpha x+\beta x)+\sin(\alpha x-\beta x))dx\\&=\lim_{t\to\infty}\frac1{2t}\left(\frac{\cos((\alpha-\beta)t)}{\alpha-\beta}-\frac{\cos((\alpha+\beta)t)}{\alpha+\beta}-2\right)\end{aligned}$

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    $\begingroup$ What have you tried? $\endgroup$ – Riemann'sPointyNose Jul 7 '20 at 20:29
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    $\begingroup$ Were you able to evaluate the integral? What happened then? $\endgroup$ – saulspatz Jul 7 '20 at 20:30
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    $\begingroup$ Please include your work in the question. It will help people give better feedback/answers $\endgroup$ – QC_QAOA Jul 7 '20 at 20:33
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    $\begingroup$ Your solution is indeed correct. $\endgroup$ – marty cohen Jul 7 '20 at 21:06
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    $\begingroup$ I think the person who voted to close the question before editing (both the vote and the edit after were justified at that moment) to retract the close vote because there is no need for this question to be in the close queue after all since the OP included his/her own work/attempts. $\endgroup$ – Invisible Jul 7 '20 at 23:21
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There is indeed a simple way. Express each sine and cosine in terms of exponentials via Euler formula. Multiply things together. The "infinite time average" of exponentials is straightforward. Make sure you pay attention to all the cases (e.g. $\alpha=\pm \beta$).

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    $\begingroup$ thank you so much $\endgroup$ – xlulu Jul 7 '20 at 20:45
  • $\begingroup$ The cosines work just as well with no need for complex values. $\endgroup$ – marty cohen Jul 7 '20 at 21:07
  • $\begingroup$ Can you provide the link to (the specific) Euler formula? $\endgroup$ – UmbQbify Jul 7 '20 at 21:23
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    $\begingroup$ @UmbQbify-Key20- it's $\cos(x)=(e^{ix} +e^{-i x})/2$ and so on. $\endgroup$ – lcv Jul 7 '20 at 22:01
  • $\begingroup$ Oh that. Got it. $\endgroup$ – UmbQbify Jul 8 '20 at 0:02

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