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How do I find the following integral by converting it into a complex integral and then using residue theorem?

$$\int_0^{2\pi}e^{a \cos{\theta}}\cos({\sin{\theta}})\,d\theta$$

My approach is as follows. Substitute $z=e^{i\theta}$ so that $\cos{\theta}=\frac{z+z^{-1}}{2}$, and similarly for sine. For the cos outside the exponent, use this substitution again. But it is just giving me a string of exponents, which i don't know how to integrate. Should I use integration by parts? Any ideas?

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Assuming you meant $e^{a \cos \theta} \cos(a \sin \theta)$, notice that $\text{Re} \ e^{ae^{i \theta}} = e^{a \cos \theta} \cos(a \sin \theta)$ –  Random Variable Apr 28 '13 at 8:03
    
I added an a in your problem. Is this what you wanted? Otherwise the problem is too hard. –  TCL Apr 28 '13 at 13:47
1  
In the original source that I am using, there is NO a in the cos. There is only one i.e. in the exponent. –  ramanujan_dirac Apr 28 '13 at 13:50
    
Sorry. Then you can change it back. –  TCL Apr 28 '13 at 13:59
    
@TCL: the problem isn't that hard - see my solution below. –  Ron Gordon Apr 29 '13 at 1:53

2 Answers 2

The integral is the real part of

$$\int_0^{2 \pi} d\theta\: e^{a (cos{\theta} + i \sin{\theta})}$$

which, when you set $z=e^{i \theta}$, $d\theta = -i dz/z$, you get

$$-i \oint_{|z|=1} dz \, \frac{e^{a z}}{z} $$

which, by the residue theorem is $-i (i 2 \pi) = 2 \pi$. Therefore, the value of the original integral is $2 \pi$.

ADDENDUM

In the case that the integral is actually

$$\int_0^{2 \pi} d\theta \, e^{a \cos{\theta}} \cos{(\sin{\theta})}$$

which is the real part of

$$\int_0^{2 \pi} d\theta \, e^{a \cos{\theta}} e^{i \sin{\theta}}$$

To write as a complex integral, make the same substitution as above and, after a little manipulation, get

$$-i \oint_{|z|=1} \frac{dz}{z} \exp{\left [\frac12 (a+1) z + \frac12 (a-1) z^{-1} \right ]}$$

There is an essential singularity at $z=0$; the way to find the residue there is to expand the exponential into a Taylor/Laurent series and find the coefficient of $z^0$ (because there is a factor of $z^{-1}$ inside the integral already:

$$-i \oint_{|z|=1} \frac{dz}{z} \sum_{k=0}^{\infty} \frac{1}{2^k \, k!} \left [ (a+1) z + (a-1) z^{-1} \right ]^k $$

Using the binomial theorem, we write the sum as

$$\sum_{k=0}^{\infty} \frac{1}{2^k \, k!} \sum_{m=0}^k \binom{k}{m} (a+1)^m (a-1)^{k-m} z^{2 m-k} $$

We get the coefficient of $z^0$ in the sum, for each $k$, when $m=k/2$. This only works for even values of $k$. The result is that the coefficient of $z^0$ is, when $|a|>1$:

$$\sum_{k=0}^{\infty} \frac{\left (a^2-1\right )^k}{2^{2 k} (k!)^2} = I_0 \left(\sqrt{a^2-1}\right)$$

The integral is the real part of, by the residue theorem, $i 2 \pi$ times the residue of the essential singularity at $z=0$. Therefore

$$\int_0^{2 \pi} d\theta \, e^{a \cos{\theta}} \cos{(\sin{\theta})} = 2 \pi I_0 \left(\sqrt{a^2-1}\right)$$

where $I_0$ is the modified Bessel function of the first kind of order zero. Note that when $|a|<1$, the integral is equal to $2 \pi J_0\left(\sqrt{1-a^2}\right)$, where $J_0$ is the Bessel function of the first kind of order zero.

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Souldn't it be $e^{az}$ at the numerator? Given your variable change. –  1015 Apr 28 '13 at 13:57
    
@RonGordon: Thanks for the answer, but TCL had made the wrong edit, the original question doesn't have a $a$ in the cos, or maybe it is just a typo. –  ramanujan_dirac Apr 28 '13 at 22:04
    
@ramanujan_dirac: OK, I addressed this case in my new edit. It is quite a bit more involved. For some bizarre reason, Mathematica would not produce the general result, but I have verified it numerically. –  Ron Gordon Apr 29 '13 at 1:33
    
@RonGordon: I know $J_0$ is the Bessel function of order 0. What is $I_0$? Does it have a name? –  TCL Apr 29 '13 at 2:37
    
@TCL: Modified Bessel of 1st kind of order zero. –  Ron Gordon Apr 29 '13 at 3:14

Follow Random Variable comment, and let $z=ae^{I\theta}$. Then $dz=iz\,d\theta$ and your integral becomes $$\int_C\frac{e^z}{iz}\,dz$$ where $C$ is the circle $|z|=a$. Then you should know what to do from here. (For some reason, I cannot type in a lower letter I in the exponent of e.)

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