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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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When you write out function names like $\operatorname{sin}$ as words, $\TeX$ interprets them as a juxtaposition of variable names, which get italicized and don't have the right spacing. To get proper formatting for function names, you need to use the commands predefined for that purpose, e.g. \sin, or, for functions for which there's no predefined command, you can use \operatorname{name} to produce "$\operatorname{name}$". –  joriki Apr 19 '12 at 2:35
    
@joriki:thanks I will fix it –  Hassan Muhammad Apr 19 '12 at 2:38
    
In that answer the integral with sin is from $-\pi$ to $\pi$. Since $\sin(t)$ is odd, the integral is obviously 0... –  N. S. Apr 19 '12 at 2:47
    
You only fixed one out of the four incorrectly formatted instances of $\sin$ in the post. –  joriki Apr 19 '12 at 2:51
2  
Hassan, this is very frustrating. Please take more care in formatting your posts and in responding to comments. I've now twice pointed out the incorrect formatting of the sine functions; each time you've only corrected a single one of the four instances in the post, the second time even after I had pointed out that there are four and you had agreed and announced that you would fix them. The title is still just as badly formatted as it was originally. P.S.: I see that Arturo has fixed it now. –  joriki Apr 19 '12 at 2:57

2 Answers 2

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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If that is the case then this proof is wrong [math.stackexchange.com/questions/124868/…, robjohn answer. –  Hassan Muhammad Apr 19 '12 at 2:45
    
@HassanMuhammad Did you see the end points of integration when the sin integral vanished? They are not the same as yours ;) –  N. S. Apr 19 '12 at 2:48
    
@joriki: I will take my time before accepting your answer Mr. Joriki. –  Hassan Muhammad Apr 19 '12 at 3:15

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....

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