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I am trying to show that $\int_{-\pi}^{\pi}e^{\alpha \cos t}\sin(\alpha \sin t)dt=0$

Where $\alpha$ is a real constant.


I found the problem while studying a particular question in this room,this one. It becomes so challenging to me as I am trying to make life easy but I stucked!

EDIT: The integral is from $-\pi$ to $\pi$

EDIT 2: I am sorry for this edit, but it is a typo problem and I fix it now. In my question I have $e^\alpha \cos t$ not $e^\alpha$ only. I am very sorry.

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    Well then say that right away and I won't spend my time trying to help you with the formatting. What's the point of announcing that you'll change it and then not changing it?2012-04-19

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This is false. In the interior of the interval of integration, the value of the inner sine is in $(0,1]$. For sufficiently small $\alpha$, that means the value of the outer sine is positive, so since $\mathrm e^\alpha$ is also positive, the integral is positive.

[Edit in response to the change in the question:]

As N.S. has already pointed out, the new integral vanishes because the integrand is odd and the integration interval is symmetric about $0$. By the way, also note that $\mathrm e^\alpha$ is a non-zero constant that doesn't affect whether the integral is zero.

[Edit in response to yet another change in the question:]

The integrand is still odd; the cosine in the exponent doesn't change that. And please take more care in posting; it's a huge waste of everyone's time to ask two questions that you didn't mean to ask and have people spend time answering them.

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    @HassanMuhammad Did you see the end points of integration when the sin integral vanished? They are not the same as yours ;)2012-04-19
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Hint Your function is ODD....

This is one of the pretty standard result in Calculus:

If $f(t)$ is a continuous, odd function, then

$\int_{-a}^a f(t) dt =0$

Proof: Substitute $u=-t$.

Second Proof: Think of the integral as a signed area....