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show that

$$ \int\limits_0^{2\pi }{\frac{{{e^{\cos x}}\cos\left({\sin x}\right)}}{{p-\cos\left({y-x}\right)}}}dx =\frac{{2\pi }}{{\sqrt{{p^2}-1}}}\exp\left({\frac{{\cos y}}{{p+\sqrt{{p^2}-1}}}}\right)\cos\left({\frac{{\sin y}}{{p+\sqrt{{p^2}-1}}}}\right);\left({p > 1}\right) $$

I think this is nice integral,But I can't show it,Thank you

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  • $\begingroup$ The Maple code $$int(exp(cos(x))*cos(sin(x))/(p-cos(y-x)), x = 0 .. 2*Pi)\,assuming\, p>1, y>0 $$ produces the output which can be seen here. $\endgroup$
    – user64494
    Sep 24, 2013 at 4:44
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    $\begingroup$ $ \displaystyle e^{\cos x} \cos (\sin x) = \text{Re} \ e^{e^{ix}} $ $\endgroup$ Sep 24, 2013 at 5:24
  • $\begingroup$ @Pedro Tamaroff: Why did you delete the answer? $\endgroup$ Dec 26, 2014 at 3:06

1 Answer 1

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Observe that the numerator is simply the real part of $e^{e^{i x}}$. Thus, the desired integral is simply the real part of the contour integral

$$2 i \oint_{|z|=1} dz \frac{e^z}{e^{-i y} z^2-2 p z + e^{i y}}$$

(This is derived by substituting $z=e^{i x}$, $dx = -i dz/z$, $\cos{x}=(z+z^{-1})/2$, $\sin{x}=(z-z^{-1})/(2 i)$, and doing a little algebra.)

The poles of the integrand are at $z_{\pm}=(p \pm \sqrt{p^2-1}) e^{i y}$, of which only $z_-$ is inside the unit circle (recall that $p \gt 1$). The residue at this pole is simply

$$2 i\frac{e^{(p-\sqrt{p^2-1}) e^{i y}}}{-2 \sqrt{p^2-1}}$$

and the integral is therefore, by the residue theorem, $i 2 \pi$ times this residue, or

$$\frac{2 \pi}{\sqrt{p^2-1}} e^{(p-\sqrt{p^2-1}) e^{i y}}$$

We then take the real part of the above to get the sought-after integral. Thus,

$$\int_0^{2 \pi} dx \frac{e^{\cos{x}} \cos{(\sin{x})}}{p-\cos{(y-x)}} = \frac{2 \pi}{\sqrt{p^2-1}} e^{\left (p-\sqrt{p^2-1}\right ) \cos{y}} \cos{\left [\left (p-\sqrt{p^2-1}\right ) \sin{y}\right ]} $$

as was to be shown.

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  • $\begingroup$ Is this technique different from mine? It is exactly the idea in my answer. $\endgroup$ Sep 24, 2013 at 12:59
  • $\begingroup$ @MhenniBenghorbal: Big difference is that mine is useful and yours is not. Your idea, as I noted, beyond the fact that you sub $z=e^{i x}$, is wrong. $\endgroup$
    – Ron Gordon
    Sep 24, 2013 at 13:11
  • $\begingroup$ Thanks to the comment by Random variable. $\endgroup$ Sep 24, 2013 at 13:17
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    $\begingroup$ @MhenniBenghorbal: what's that supposed to mean? That I was only able to solve this problem because someone else had the same idea? I never claimed it was a work of genius. But, no, I did not give RV credit because I came up with that all by myself. You have no idea how I solve my problems, so it is quite presumptuous of you to post, on my answer, that I owe someone else credit. $\endgroup$
    – Ron Gordon
    Sep 24, 2013 at 13:24

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