How and when to use Cauchy's integral formula

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Hello.
How do I know when to use Cauchy integral formula. Why do we use the formula in this question? As you can see in my attempt, I got stuck.
Are my values for f(z), z, z​0 here correct?
 

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You got the right answer, since f(zo)=1.
You use the Cauchy integral when you have to evaluate an integral that matches the pattern of a Cauchy integral!
Sometimes, the contour needs to be closed to get a match and the integral on the closing path can be evaluated by some other means.
The Cauchy integral is very useful in physics and in signal theory.
 
maajdl said:
You got the right answer, since f(zo)=1.
You use the Cauchy integral when you have to evaluate an integral that matches the pattern of a Cauchy integral!
Sometimes, the contour needs to be closed to get a match and the integral on the closing path can be evaluated by some other means.
The Cauchy integral is very useful in physics and in signal theory.

No, I just solved it and f(z) is not 1, but 1/(z-1).
 
You proved that your integral is 2 Pi I * the Cauchy integral of f(z)=1 around z=I .
Therefore, your integral is 2 Pi I f(I) = 2 Pi I .
 
MissP.25_5 said:
Hello.
How do I know when to use Cauchy integral formula. Why do we use the formula in this question? As you can see in my attempt, I got stuck.
Are my values for f(z), z, z​0 here correct?

I never look at photo attachments of handwritten work; in fact, if you read the "PF Guidelines" post by vela, you will see that you are not supposed to use them except for very special circumstances---for several good reasons. You should take the trouble to type out your work if you want the helpers to take the trouble to offer free assistance.
 
Ray Vickson said:
I never look at photo attachments of handwritten work; in fact, if you read the "PF Guidelines" post by vela, you will see that you are not supposed to use them except for very special circumstances---for several good reasons. You should take the trouble to type out your work if you want the helpers to take the trouble to offer free assistance.

Exactly! Thread locked. Please create a thread where you write out your work.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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