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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? $\endgroup$ Jul 7, 2020 at 20:29
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    $\begingroup$ Were you able to evaluate the integral? What happened then? $\endgroup$
    – saulspatz
    Jul 7, 2020 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, 2020 at 20:33
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    $\begingroup$ Your solution is indeed correct. $\endgroup$ Jul 7, 2020 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. $\endgroup$
    – PinkyWay
    Jul 7, 2020 at 23:21

1 Answer 1

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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, 2020 at 20:45
  • $\begingroup$ The cosines work just as well with no need for complex values. $\endgroup$ Jul 7, 2020 at 21:07
  • $\begingroup$ Can you provide the link to (the specific) Euler formula? $\endgroup$
    – UmbQbify
    Jul 7, 2020 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, 2020 at 22:01
  • $\begingroup$ Oh that. Got it. $\endgroup$
    – UmbQbify
    Jul 8, 2020 at 0:02

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