# Improper integrals determine if they converge or diverge.

The question is as follows.

Determine if the these 2 improper integrals converge.

$\int^{\infty}_{0} ( x^{1/2} +x^{3/2} )^{-1}$ And $\int^{\pi}_{0} (1-\cos(x))/(\sin^{2}(x))$

For the first integral we have problems at post infinty and $0$ clearly $1$ is ok so let's split it up into

$\lim_{c\to 0} -\int^{c}_{1} ( x^{1/2} +x^{3/2} )^{-1}$ + $\lim_{d\to\infty}\int^{d}_{1} ( x^{1/2} +x^{3/2} )^{-1}$

now my problem being is I can't even evaluate these limits without integrating them clearly I could bound the first one with $x^{-2}$ when d goes to infinity as it both gets the bound much faster and secondly $x^{-(1/2)}$ as $c$ does to $0$ as well as

so we have $\lim_{c\to 0}$ $1/(c)^{1/2}$ this diverges so we are done? as for the second problem I am not sure how to approach it

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On the interval from $0$ to $1$, use the fact that the function we are integrating is $\le (x^{1/2})^{-1}$. On the interval from $1$ to $\infty$, use the fact that the function we are integrating is $\lt (x^{3/2})^{-1}$.

Since $\displaystyle\int_0^1 \dfrac{dx}{x^{1/2}}$ and $\displaystyle\int_1^\infty \dfrac{dx}{x^{3/2}}$ both converge, we are finished.

For the second problem, it turns out everything is OK at $0$, and for say the integral from $0$ to $\pi/2$. For details you might want to multiply top and bottom by $1+\cos x$.

But it is a bit of a waste of time, since there is bad trouble at $\pi$.

For the integral from $\pi/2$ to $\pi$, I suggest for the sake of familiarity letting $y=\pi -x$. You will end up with an integral that diverges, because of a $\sin^2 y$ at the bottom.

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<--- moron sorry when i get tired i stop making any sense your totally right on making the choice of what to integrate i dont know how i forgot that – Faust7 Mar 22 '13 at 6:16