improper integral: int from 0 to inf : 1 / (sqrt(x)(1+x))

[rlauren]

Hello,

I need help finding an appropriate strategy for solving the following integral. I understand that both bounds create a problem and that I need to use limits to solve it. I do not know how to go about actually figuring out the integral, however. I am not sure if I should use integration by parts or if some other strategy would work.

int from 0 to inf : 1 / (sqrt(x)(1+x))

Should I break this into 1 / sqrt(x) * 1 / (1+x) and use integration by parts? Is there some other process that would work?

Thanks!!!

 

[galactus]

Is this your integral:

\(\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x}(1+x)}dx\)

If so, we can integrate by making a sub as so:

Let \(\displaystyle u=\sqrt{x}, \;\ x=u^{2}, \;\ 2udu=dx\)

Making the subs gives:

\(\displaystyle 2\int_{0}^{\infty}\frac{1}{1+u^{2}}du\)

Now, see what this integral is?. It is a rather popular one.

HINT: think about arctan

 

[rlauren]

ahh... ok,

so then I end up with the following... If I break the bounds up to have from 0 to 1 and 1 to inf with a approaching 0 and b approaching inf:

lim a -> 0 [ 2 tan^-1(sqrt(1)) - 2 tan^-1 (sqrt(a))] + lim b -> inf [ 2 tan^-1(sqrt(b)) - 2 tan^-1 (sqrt(1))] =

2 [tan^-1(1) - tan^-1(0)] + 2[tan^-1(inf) - tan^-1(1)] =

2 tan^-1(1) + 2[(pi/2) - tan^-1(1)] is this correct???

 

[galactus]

It is hard to follow all that and I have to leave soon, but \(\displaystyle 2tan^{-1}(\sqrt{x})\) is correct.

Now, just use the limits:

\(\displaystyle 2tan^{-1}(\sqrt{\infty})-2tan^{-1}(\sqrt{0})={\pi}\)

 

[rlauren]

So I do not need to create new bounds ? What I mean is, in the past, when I have had a problem where the upper and lower bounds both cause problems I would instead add together two integrals with different bounds.

for instance, here it would be the integral from a (as a approaches 0) to 1 plus the integral from 1 to b (as b approaches inf)?


Nevermind, I think it ends up with the same answer. Thanks!

 

[DrMike]

So I do not need to create new bounds ? What I mean is, in the past, when I have had a problem where the upper and lower bounds both cause problems I would instead add together two integrals with different bounds.

for instance, here it would be the integral from a (as a approaches 0) to 1 plus the integral from 1 to b (as b approaches inf)?

Nevermind, I think it ends up with the same answer. Thanks!

This approach is fine, and technically something like this is necessary for this problem, since the integrand goes to infinity as \(\displaystyle x\rightarrow 0^+\). However, the lazy way (treating integrals from 0 as proper instead of improper) works fine for this problem.

An alternative to splitting the integral is to say

\(\displaystyle \int_0^\infty\frac{1}{\sqrt{x}(1+x)}dx=\lim_{U\rightarrow\infty,L\rightarrow0^+}\int_L^U\frac{1}{\sqrt{x}(1+x)}dx=\lim_{U\rightarrow\infty,L\rightarrow0^+}\int_{\sqrt{L}}^{\sqrt{U}}\frac{1}{1+u^2}du=\lim_{U\rightarrow\infty}\int_0^{\sqrt{U}}\frac{1}{1+u^2}du\)