Using Cauchy's integral formula to evaluate integrals

Woolyabyss
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Homework Statement


Use Cauchy’s integral formula to evaluate the integral along γ(t) of (z/(z+9)^2)dz
where γ(t) = 2i + 4e^it , 0 ≤ t ≤ 2π.

Homework Equations


Cauchy's integral formula

The Attempt at a Solution


I was just wondering is the integral not just zero by Cauchy's theorem since (z/(z+9)^2) is holomorphic inside the circle defined by γ(t) ( the singularity at -9 is outside the circle ).
 
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Woolyabyss said:

Homework Statement


Use Cauchy’s integral formula to evaluate the integral along γ(t) of (z/(z+9)^2)dz
where γ(t) = 2i + 4e^it , 0 ≤ t ≤ 2π.

Homework Equations


Cauchy's integral formula

The Attempt at a Solution


I was just wondering is the integral not just zero by Cauchy's theorem since (z/(z+9)^2) is holomorphic inside the circle defined by γ(t) ( the singularity at -9 is outside the circle ).
Yes, that seems to be the case.
 
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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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