Can you me proving , uniform convergence implies pointwise convergence

shehpar
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I don't know, I started from the definition of uniform convergence and it seems pretty obvious to me , can anybody start me at least towards right direction?
 
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Yes, that pretty much all it takes.

Definition of "uniformly convergent": \{f_n(x)\} converges to f(x) as n goes to infinity uniformly if and only if, given \epsilon> 0, there exist N such that if n > N, then |f_n(x)- f(x)|< \epsilon.

Definition of "convergent": {f_n(x)} converges to f(a) for given a as n goes to infinity if and only if given \epsilon> 0, there exsit N such that if n> N, then |f_n(a)- f(a)|< \epsilon.

Basically, "uniformly convergent" requires that, given \epsilon, you be able to use the same \delta for every value of x. Just "convergent" means the value of \delta may depend upon \epsilon and the value of x at which the function is evaluated.
 
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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