Problem with Improper Integration - Is Zero the Correct Answer?

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Hi again,

I have a problem regarding improper integration.

Homework Statement

feafea.JPG


Refer to the image. I tried to solve and got zero for the answer. Is that correct? I refer to my actual problem it seem like it don't won't this way...

Thanks
 
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suppose that g(t)=1, what is the value of the integral?
 
benorin said:
suppose that g(t)=1, what is the value of the integral?

If 1 is integrated, we get t and then sub in the boundary value the answer still 0?

here the actual g(t) suppose to be g(t) = cos(2\pif0t) - jsin(2\pif0t).
 
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killerfish said:
If 1 is integrated, we get t and then sub in the boundary value the answer still 0?
Are you saying that you are not capable of integrating
\int_{-T/2}^{T/2} dt
or you just haven't bothered to?

here the actual g(t) suppose to be g(t) = cos(2\pif0t) - jsin(2\pif0t).
 
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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