Prove that the integral diverge.

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Let f:[0,inf) \rightarrow R+ (f is positive...).
f is continuous and: [PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP350919g712a61ice7c02000046b8i2i10g466041?MSPStoreType=image/gif&s=15&w=87&h=35.

prove that :[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP334719g7149aa8g6efha000027bhe3a6fe5cg8c9?MSPStoreType=image/gif&s=22&w=93&h=40

where [PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP165019g717ia4c72gibe000039f8g9b652gd3664?MSPStoreType=image/gif&s=41&w=99&h=35.

some tips?
thanks.
 
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You say that \int_0^\infty f(x) dx= \infty and then g(x)= \int_0^\infty f(t)dt. Did you mean g(x)= \int_0^x f(t)dt?
 
Yes, sorry for the mistake, this is exactly as you mentioned.
 
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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