I need help in solving the following problem in complex calculus. I need to calculate this real integral: $$ \int_0^{2\pi} e^{\cos{\theta}}\cos(\sin\theta) \,\,\, d\theta $$

I think that the best method is to do the substitution $z=re^{i\theta}$, from which we have $d\theta = \frac{dz}{iz}$, $\cos\theta=\frac{1}{2}(z+\frac{1}{z})$ and so on. The integral then becomes (without constants) $$ \oint_{|z|=r} e^{\frac{z}{2}}e^{\frac{1}{2z}}\frac{z^4-6z^2+1}{z(z^2-1)} \,\,\,dz $$

At this point I am unable to calculate the residue at the essential singularity in 0, because the manipulation of Laurent series rapidly becomes utterly complicated and I don't see any simple way to find the final coefficient of $1/z$. So, which is the way to handle this integral? Another substitution?

  • $\begingroup$ It seems like it's impossible to expand Exp(1/x) into Laurent series, because it has infinite number of negative degrees of x. I mean, for every n: Exp(1/x) * x^n diverges at x=0. $\endgroup$ – valdo Sep 18 '11 at 14:59
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    $\begingroup$ Here's an idea: the integral of $e^{\cos{\theta}}\sin(\sin\theta)$ is zero over the same interval (why?). Add $i$ times that integrand to your original integrand, and see if anything simplifies... $\endgroup$ – J. M. isn't a mathematician Sep 18 '11 at 15:03

Things work very nicely if we take a slightly different approach at the start. (The value of the integral is $2\pi$.) Notice that $$\cos\left(\sin\theta\right)=\text{Re}\left(e^{i\sin\theta}\right).$$ Hence, since $e^{\cos\theta}$ is already real our integral becomes $$\int_{0}^{2\pi}\text{Re}\left(e^{\cos\theta}e^{i\sin\theta}\right)d\theta=\text{Re}\int_{0}^{2\pi}e^{\left(e^{i\theta}\right)}d\theta.$$

Substituting $z=e^{i\theta}$ we have $$\text{Re}\int_{\mathbb{S}^{1}}\frac{e^{z}}{iz}dz=2\pi i \text{Res}\left(\frac{e^{z}}{iz}\right) = 2\pi$$ since the residue is $\frac{1}{i}$.

Hope that helps,

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  • $\begingroup$ We have done the calculation together in this minutes... I've used the suggestion given above by J.M. but even your solution is valuable and very smart. Thank you! $\endgroup$ – 5olidu5 Sep 18 '11 at 16:02

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