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Let $$f(z)=\frac{e^{\pi iz}}{z^2-2z+2}$$ and $\gamma_R$ is the closed contour made up by the semi-circular contour $\sigma_1$ given by, $\sigma_1(t)=Re^{it}$, and the straight line $\gamma_2$ from $-R$ to $R$ (a semi circle).

So the singularities of $f$ are $1+i$ and $1-i$ and it's type is 'double'.

The residue theorem is $$\oint f(z)dz=2\pi i\sum_{k=1}^n I(\gamma,a_k)Res(f,a_k).$$

I'm not sure how to use this theorem to calculate the residues though.

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Would I be able to use the answer to this to evaluate $$\int_{-\infty}^\infty\frac{\cos(\pi x)}{x^2-2x+2}dx$$ and $$\int_{-\infty}^\infty\frac{\sin(\pi x)}{x^2-2x+2}dx?$$ – Levi Jan 18 '13 at 0:42
Yes. Look at the real and imaginary parts of the original integral separately. – mrf Jan 18 '13 at 6:19

The residue of the function inside the contour is

$$\lim_{z \rightarrow 1+i} (z-(1+i)) f(z) $$

$$ = \frac{e^{i \pi (1+i)}}{2 i} $$

The value of the integral is simply $i 2 \pi $ times this residue.

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Would multiplying $\frac{e^{i\pi}}{2i}$ by $i2\pi$ give us the answer to $\int_{\gamma R} f(z) dz$ for $R>2$? – Levi Jan 18 '13 at 0:48
No. The answer is $i 2 \pi \frac{e^{i \pi} e^{-\pi}}{2 i} = -\pi e^{-\pi}$. This holds for $R > \sqrt{2}$. – Ron Gordon Jan 18 '13 at 1:15

Each pole is a simple pole (not double). The easiest way to compute the residue is

$$\operatorname{Res}(f,1+i) = \frac{e^{\pi i z}}{(z^2-2z+2)'}\bigg|_{z=1+i} = \frac{e^{\pi i(1+i)}}{2(1+i)-2} $$

with a similar formula for the other pole (but the other one is not needed for your integral).

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The residue in fact equals simply $\,-\frac{e^{-\pi}}{2i}=\frac{e^{-\pi}i}{2}\,$ – DonAntonio Jan 18 '13 at 4:00
Of course. I left some algebra for the OP. – mrf Jan 18 '13 at 6:11

When the numerator is an entire function, the residue at each zero of the denominator is the same as if you just had a constant over the denominator, and the constant is the value of the numerator at that point.

(....or even if the numerator is merely differentiable in some open neighborhood of that point.)

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