Evaluating Integrals using the Residue THM

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Homework Statement



integral |z|=1 of sinz/z2dz


Homework Equations


Rule #1 if f(z) has a simple pole at z0, then
Res[f(z),z0] = lim(as z goes to z0) (z - z0)*f(z)

Rule #2 if f(z) has a double pole at z0, then
Res[f(z),z0] = lim(as z goes to z0)d/dz (z - z0)2*f(z)

Rule #3 If f(z) and g(z) are analytic at z0, and if g(z) has a simple zero at z0 then,
Res[f(z)/g(z), z0] = f(z0)/g'(z0)

Rule #4 If g(z) is analytic and has a simple zero at z0 then,
Res[1/g(z), z0] = 1/g'(z0)





The Attempt at a Solution



Just confused on which one to use? Would I integrate first? And if so what would my z0 be in Rest[..., z0]

Thanks
 
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What kind of of pole does the function sinz/z2 have and where is it? Although I think 1 and 2 will give the same result.
 
phsopher said:
What kind of of pole does the function sinz/z2 have and where is it? Although I think 1 and 2 will give the same result.

double pole at i/2 and -i/2?

Edit: Actually simple pole at z = 0?
 
Last edited:
Yep.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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