Finding Whether Improper Integrals Converge

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Hi all,

I'm having trouble with finding an improper integral.

The problem is ∫10(xln(x))dx

My book says the answer is -1/4, but I do not understand how this is the case.

lim(xlnx) as x approaches 1- = 0

lim(xlnx) as x approaches 0+ = ∞

So how does this value converge at -1/4?

Thanks in advance!
 
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rocapp said:
lim(xlnx) as x approaches 0+ = ∞

What makes you say that? Did you try Lhopital?
 
Ah. I see that. So now I just find the integral and from zero to one, correct?
 

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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