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The contour integral is:
$$\int_{\Gamma}\frac{\cos z + i \sin z}{(z^2 + 36)(z+2)} \mathrm{d} z$$
where $\Gamma$ is the circle centred at the origin, with radius 3, traversed once positively.

I computed this contour integral to be $\frac{\pi i e^{-2}}{20}$, however, WolframAlpha is claiming that it's equal to zero when I try to do it by parametrisation with $z:= e^{it}$ for $0 \leq t \leq 2\pi$.
W|A Link: http://www.wolframalpha.com/input/?i=integral+from+0+to+2pi+of+(e%5E(i(e%5E(it)))%2F((e%5E(2it)+%2B+36)(e%5E(it)%2B2))+++e%5E(it)++i+dt

Why does W|A give a different answer? Am I wrong then with my original answer?

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3 Answers 3

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You gave Wolfram Alpha the wrong integral to do. Your parametrization was $z = e^{it}$, but this gives a circle of radius $1$, not a circle of radius $3$. With the correct parametrization $z = 3 e^{it}$, Wolfram Alpha gives the correct answer.

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Let $f(z)=\frac{e^{iz}}{z^2+36}$. Then, by the integral formula:

$$\int_\Gamma\frac{e^{iz}}{(z+2)(z^2+36)}\,dz=\int_\Gamma\frac{f(z)}{z+2}\,dz= 2 \pi i f(-2)=\frac{\pi i e^{-2i}}{20}.$$

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This is $$\int_\Gamma\frac{e^{iz}}{(z+2)(z^2+36)}\,dz.$$ The only pole inside $\Gamma$ is at $z=-2$ and is simple. The residue is $$\frac{e^{-2i}}{40}$$ so the integral is $$\frac{\pi i e^{-2i}}{20}.$$

As for why Wolfie A gives zero, that I suspect will be forever a mystery.

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