Improper Integrals: Struggling to Understand Convergence

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Homework Statement



I don't understand why the following intergral is improper
http://img413.imageshack.us/img413/1667/38410700.jpg

I have a graph for the function.
http://img413.imageshack.us/img413/2424/photoxxs.jpg

According to this graph, the function should converge in [-∞,0)
However, my solution manual says that the function has an infinite interval of integration.

Could anyone tell me what I am mistaken?

Thank you for reading.
 
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Improper integrals may either converge or diverge, that is, have a finite value or not. An improper integral is an integral involving an infinity in either or both bounds, or one whose integrand is discontinuous over the interval.
 
Oh I should have studied more.

Thank you. :)
 
"Infinite interval of integration" simply means that you are integrating from -infinity to 0, an infinite length.
 

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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