# Thread: Improper integrals of type I: convergent or divergent?

1. ## Improper integrals of type I: convergent or divergent?

I've got two improper integrals, and need to determine whether or not they are convergent.

1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

2) (x^2)e^(-x^2) from 0 to infinity
No idea where to go with this one, any hints? Something I should look for?

Thanks!!

2. Originally Posted by mistykz
I've got two improper integrals, and need to determine whether or not they are convergent.

1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

Thanks!!
Hint)

$\displaystyle{\int\limits_1^\infty\frac{dx}{x\sqrt {x+3}}<\int\limits_1^\infty\frac{dx}{x\sqrt{x}}}=\ lim\limits_{b\to\infty}\int\limits_1^b\frac{dx}{x^ {3/2}}=\left.{-2\lim\limits_{b\to\infty}\frac{1}{\sqrt{x}}}\right |_1^b=-2\lim\limits_{b\to\infty}\!\left(\frac{1}{\sqrt{b} }-1\right)=2}$

3. Ah, that makes sense! Thanks!!

4. Originally Posted by mistykz
I've got two improper integrals, and need to determine whether or not they are convergent.

1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
I said that this function was =< 1/x, and was hoping to say that the integral from 1 to infinity of 1/x converged, therefore my given function would... but that's not the case Can someone point me in the right direction? I know that it's supposed to converge, but is there an easy way to see this?

2) (x^2)e^(-x^2) from 0 to infinity
No idea where to go with this one, any hints? Something I should look for?

Thanks!!
$\displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^\infty{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}} \right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\limits_ 0^b{e^{-x^2}\,dx}=}$

$\displaystyle{=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2 }}+\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^ {x^2})'}+\frac{\sqrt{\pi}}{4}=}$

$\displaystyle{=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^ 2}}+\frac{\sqrt{\pi}}{4}=-\frac{1}{4}\cdot0+\frac{\sqrt{\pi}}{4}=\frac{\sqrt {\pi}}{4}}$

See Error function - Wikipedia, the free encyclopedia

5. Thanks! Although I believe that seems to be outside the scope of our course, it definitely works lol

6. Originally Posted by mistykz
2) (x^2)e^(-x^2) from 0 to infinity
No idea where to go with this one, any hints? Something I should look for?
Thanks!!
Originally Posted by DeMath
$\displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^\infty{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}} \right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\limits_ 0^b{e^{-x^2}\,dx}=}$

$\displaystyle{=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2 }}+\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^ {x^2})'}+\frac{\sqrt{\pi}}{4}=}$

$\displaystyle{=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^ 2}}+\frac{\sqrt{\pi}}{4}=-\frac{1}{4}\cdot0+\frac{\sqrt{\pi}}{4}=\frac{\sqrt {\pi}}{4}}$
This solution is better

$\displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}=-\frac{1}{2}\int\limits_0^b{x\,d(e^{-x^2})}=\left.{-\frac{1}{2}xe^{-x^2}}\right|_{x=0}^{x\to\infty}+\frac{1}{2}\int\li mits_0^\infty{e^{-x^2}\,dx}}$

$\displaystyle{\left.{-\frac{1}{2}xe^{-x^2}}\right|_{x=0}^{x\to\infty}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{x}{e^{x^2 }}=-\frac{1}{2}\lim\limits_{x\to\infty}\frac{(x)'}{(e^ {x^2})'}=-\frac{1}{4}\lim\limits_{x\to\infty}\frac{1}{xe^{x^ 2}}=-\frac{1}{4}\cdot0=0}$

$\displaystyle{\int\limits_0^\infty{e^{-x^2}\,dx}=\int\limits_0^1{e^{-x^2}}\,dx}+\int\limits_1^\infty{e^{-x^2}\,dx}}$

The integral $\displaystyle{\int\limits_0^1{e^{-x^2}}\,dx}$ is own, that is convergent.

$\displaystyle{\int\limits_1^\infty{e^{-x^2}\,dx}<\int\limits_1^\infty\frac{dx}{x^2}=\left .{-\lim\limits_{b\to\infty}\frac{1}{x}}\right|_1^b=-\lim\limits_{b\to\infty}\!\left(\frac{1}{b}-1\right)=1}$

So, the integral $\displaystyle{\int\limits_0^\infty{x^2e^{-x^2}\,dx}}$ is convergent.

7. Originally Posted by mistykz
I've got two improper integrals, and need to determine whether or not they are convergent.

1) 1/(x*sqrt(x+3)) integrated from 1 to infinity
all you gotta do is to bound it and conclude namely that $\displaystyle\int_1^\infty\dfrac{dx}{x^n}<\infty$ if $n>1.$

Originally Posted by mistykz
2) (x^2)e^(-x^2) from 0 to infinity
play with inequalities, just analise it for $x\ge1,$ then $-x^2\le-x$ so $e^{-x^2}\le e^{-x},$ which implies that $x^2e^{-x^2}\le x^2e^{-x},$ and the one of the right side is known to converge.

8. Note that I added a line in my last post.