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I am stuck trying to show that $\displaystyle \int\limits_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^x+1}=0$

I have tried using a Squeeze Theorem type approach, but at $\pi/4$ any function I choose overlaps and becomes less than or greater than the function depending. I am not sure where to start anymore. Anything will be helpful. Thank you in advance!

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2 Answers 2

up vote 4 down vote accepted

There is a result stated as follows

Suppose $f$ is continuous and even on $[-a,a]$, $a>0$ then $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$

Using this result, $\displaystyle \int_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^x+1}dx=\int_{0}^{\pi/2}\cos(2x)\,dx=0$.

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interesting link. Could you please check if this is really what you intended to link to? –  Fabian Jan 8 '13 at 10:10
    
@pipi We can replace $e^{x}$ with $a^x$ where $a\in\mathbb{R}^{+}$. :] –  NeverBeenHere Jan 8 '13 at 12:42
    
Oh, I've realised that we've already used $a$. Say then $\alpha^x$ to avoid confusion. –  NeverBeenHere Jan 8 '13 at 12:56
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@Artist We may replace $1+e^x$ by $1\pm \alpha ^{k(x)}$ for any odd function $k(x)$. See here –  pipi Jan 9 '13 at 23:42
    
@pipi Very nice! –  NeverBeenHere Jan 10 '13 at 14:43

Let $x \mapsto -x$ then we get $\displaystyle I = \int_{-\pi/2}^{\pi/2}\frac{\cos(2x)}{e^{-x}+1}$ -- adding them we have:

$\displaystyle 2I = \int_{-\pi/2}^{\pi/2} \frac{\cos(2x)}{e^x+1}+\frac{\cos(2x)}{e^{-x}+1}\;{dx}= \int_{-\pi/2}^{\pi/2}\cos(2x) = 0$. Thus $I= 0$.

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Thank you so much! That was exactly what I was looking for. –  SyntacticSugar Jan 8 '13 at 8:13
    
@AmrDiab Glad I was of help. :] –  NeverBeenHere Jan 8 '13 at 8:18
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This is probably a more fundamental question, but why can you let x be -x for one part only? –  SyntacticSugar Jan 8 '13 at 8:29
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@AmrDiab Not sure what you're asking but $\cos(-2x) = \cos(2x)$ as the cosine function is even and the limits get inverted back to their back to their original positions by the negative from the differential. If you're confused by $x \mapsto -x$ then let $x = -t$ then change $t$ to $x$ (dummy variable) when adding. –  NeverBeenHere Jan 8 '13 at 8:35
    
Ohh okay. I wasn't sure if you could use a dummy variable like that. Thank you again. –  SyntacticSugar Jan 8 '13 at 8:37

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