Solving the Integral in Proving Analyticity of g(z)

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Homework Statement



My textbook has this question

f is continuous on C which is a simple closed contour...

Prove g(z) = 1/(2*pi*i) int_c f(s)ds/(s-z) is analytic?

I understand that you use Cauchy-Riemann proof for analytic functions, but how do I find the derivatve of a function that has an integral inside of it? Any suggestions?



Homework Equations



I thought maybe somehow I'd have to use the Cauchy Integral formula, but what version of it should I use? and should I even use that - the only reason I think that is that is the question in the textbook after the Cauchy Integral topic! :)

The Attempt at a Solution



I'm not sure where to start because I don't know how to differentiate the funciton because of the integral there and not sure how to get rid of it!
 
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laura_a said:

f is continuous on C which is a simple closed contour...
Prove g(z) = 1/(2*pi*i) int_c f(s)ds/(s-z) is analytic?


Is that exactly how the question is stated? Does it say that f is continuous on C, or does it say f is analytic on and inside C?
 
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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