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Let $I=\displaystyle\int_{0}^{+\infty}\dfrac{dx}{x^2+x^a}$.

I need to find all $a$ such that $I$ converges.

Please help me!

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One, why so hectic? Two, you show no work and it looks like you are assigning us homework. Three, this is generally not useful to anyone, including yourself, unless you give us something. -1 – Jebruho Dec 18 '12 at 6:25
What happens with this question of yours? – Did Dec 18 '12 at 6:57

There are two potential sources of badness: the length of the interval, and the function blowing up badly near $0$. So let us separate our integral into two parts: If either one is bad, the whole thing is bad. If both are good, the whole thing is good. It doesn't really matter, but lets break things up as the integral from $0$ to $1$, and the integral from $1$ to $\infty$.

1 to infinity: There is no problem, our function is $\lt \dfrac{1}{x^2}$, and it is a standard fact that $\displaystyle\int_1^\infty \dfrac{dx}{x^2}$ converges. You could write out a detailed argument by evaluating this integral from $1$ to $M$, and letting $M\to\infty$.

0 to 1: Recall that $\displaystyle\int_0^1 \dfrac{dx}{x^p}$ diverges if $p\ge 1$, and converges if $p\lt 1$. Let's examine our integral at $a=1$. Note that $x^2\lt x$, and therefore $$\frac{1}{x^2+x}\gt \frac{1}{2}\cdot\frac{1}{x}.$$ Since $\displaystyle \int_0^1\dfrac{dx}{x}$ diverges, so does $\displaystyle \int_0^1\dfrac{dx}{x^2+x}$.

Your turn: show that $\displaystyle \int_0^1\dfrac{dx}{x^2+x^\alpha}$ also diverges if $a\gt 1$.

Finally, we want to show that $\displaystyle \int_0^1\dfrac{dx}{x^2+x^\alpha}$ converges if $a\lt 1$. Use the fact that $x^2+x^a\gt x^a$.

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Do you know how to calculate $I=\displaystyle\int_{0}^{t}\dfrac{dx}{x^2+x^a}$?

If not, do a=1 and t=1, and after try to generalize.

Then, after take the limit $\lim_{t\to +\infty}\displaystyle\int_{0}^{t}\dfrac{dx}{x^2+x^a}$ and see when it converges.

Any questions I'm here.

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Some care is in place here as if $\,\alpha>0\,$ then the integrand function isn't bounded in some (right) neighbourhood of zero, so another limit of the form $\,\epsilon\to 0\,$ must be used there – DonAntonio Dec 18 '12 at 11:41

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