Convergence of Improper Integral: Can a Comparison Test Be Used?

aostraff
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Homework Statement


I'm trying to show that this improper integral converges
\int_{0}^{1} \sin \left ( x + \frac{1}{x} \right )dx

Homework Equations


The Attempt at a Solution


I thought a comparison test would be nice but I can't think of the right one if that is the way to go. I don't think a substitution of u = x + 1/x works too well as well.

Thanks.
 
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I see. Thanks.
 
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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