# Help With $\int\nolimits_{-\pi}^\pi{\cos{(y e^{i\cdot 5\cdot t} + z e^{i\cdot 3\cdot t})}dt}$

I have a couple questions that I'd like answered:

(1) How do I evaluate this integral?

(2) Is it possible to do a substitution? What about $u=e^{i \cdot t}$?

(3) What if 5 and 3 are replaced by other constant integers?

(4) What should be the most effective method to handle question (3), using the fewest calculations (like in numerical methods)? In other words, what is the most effecient way to handle the general case? (Note that it doesn't have to be a numerical method; I just want something fast!)

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I'm wondering if a substitution like $u=\cos{t} + i \sin{t}$ would work... – Matt Groff Nov 29 '10 at 17:38
Say that $y$ and $z$ are naturals, to help. – Matt Groff Nov 29 '10 at 21:12

This is another contour integral. Replace $y$ by $a$ and $z$ by $b$ to simplify notation. Then your integral is $$-i \int_{-\pi}^{\pi}{\cos(ae^{5it} + be^{3it}) \over e^{it}} ie^{it}\,dt$$ $$= -i \int_{|z| = 1}{ \cos(az^5 + bz^3) \over z}\,dz$$ By the residue theorem this integral is $2\pi$ times the residue of $\cos(az^5 + bz^3)$ at $z = 0$. Since $\cos(az^5 + bz^3) = 1$ when $z = 0$, this residue is $1$. So your overall integral is $2\pi$.
Note that it doesn't matter what $a$ or $b$ are, and that $3$ and $5$ can be replaced by any nonnegative integers.
@Zaricuse: So I'm wondering, If I replace the expression inside your cosine by a complicated expression of variables and $z$, all I have to do to find the integral value is find what limit this approaches as $z$ approaches zero? In other words, If I replace $az^5 + bz^3$ by $az^5 + bz^3 + c/z$, could I just take the limit (of this new expression) as $z$ approaches zero? – Matt Groff Nov 30 '10 at 0:45
Well, if you replace it by $az^5 + bz^3 + c/z$ you would look at the Laurent expansion of ${\cos(az^5 + bz^3 + c/z) \over z}$ around $z = 0$, identify the coeffficient $d_{-1}$ of the ${1/z}$ term, and your answer is $2\pi d_{-1}$. The same works for any $\cos(f(z))$ where $f(z)$ is analytic on some punctured disk $\{z: 1 + \epsilon > |z| > 0\}$, because such functions have Laurent expansions. – Zarrax Nov 30 '10 at 0:53
Typo correction: it's $2\pi i d_{-1}$. – Zarrax Nov 30 '10 at 1:00