I'm not sure where to begin with this problem. I've tried splitting the integral and u-substitution but I get lost either way.

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- Jul 30th 2012, 03:02 PMjeduhihelp solving integral_0^infinity 1/(sqrt(x)+1) dx
I'm not sure where to begin with this problem. I've tried splitting the integral and u-substitution but I get lost either way.

- Jul 30th 2012, 03:19 PMPlatoRe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
Look at the page

Click the show steps tab. - Jul 30th 2012, 03:54 PMjeduhiRe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
- Jul 30th 2012, 04:16 PMPlatoRe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
The integral does not converge. Look at this.

- Jul 30th 2012, 05:14 PMjeduhiRe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
Sorry I wasn't specific in my post, but I need to give reason of why it doesn't converge

- Jul 30th 2012, 08:02 PMProve ItRe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
$\displaystyle \displaystyle \begin{align*} \int_0^{\infty}{\frac{1}{\sqrt{x} + 1}\,dx} &= \int_0^{\infty}{\frac{2\sqrt{x}}{2\sqrt{x}\left( \sqrt{x} + 1 \right)} \,dx} \\ &= \int_1^{\infty}{\frac{2(u - 1)}{u}\,du} \textrm{ after making the substitution } u = \sqrt{x} + 1 \implies du = \frac{1}{2\sqrt{x}}\,dx \\ &= \int_1^{\infty}{2 - \frac{2}{u}\,du} \\ &= \left[ 2u - 2\ln{|u|} \right]_0^{\infty} \\ &= \lim_{\beta \to \infty}\left[ 2\beta - 2\ln{|\beta|} \right] - \lim_{\alpha \to 0} \left[ 2\alpha - 2\ln{|\alpha|}\right] \end{align*}$

It should be pretty obvious from the limits obtained that the integral does not converge. - Aug 1st 2012, 07:34 PMGJARe: help solving integral_0^infinity 1/(sqrt(x)+1) dx
Hi, jeduhi. I thought of another approach you might take with your exercise. The reasoning goes something like this:

1. We have some nice theorems at out disposal regarding the convergence of infinite series. Moreover, these theorems don't require much computation. Therefore, if we can recast the problem in an equivalent way using series we could save ourselves from the computational difficulties that are inherent in computing integrals using the Fundamental Theorem of Calculus (i.e. using antiderivatives).

2. There is a connection between improper integrals and series via the Integral Test for Convergence. According to this theorem, $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x}+1}$ is convergent if and only if $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}+1}$ is convergent - see Integral test for convergence - Wikipedia, the free encyclopedia for the statement of the aforementioned convergence test.

Important Note: Before using ANY theorem we must be sure that the hypothesis of the theorem are satisfied. In order to use the Integral Test for Convergence we must make sure that our function $\displaystyle f(x)=\frac{1}{\sqrt{x}+1}$ is nonnegative and strictly decreasing. By looking at the graph of this function we see that it sits above the x-axis (meaning it's nonnegative) and going down (meaning it's monotone decreasing). The fact that $\displaystyle f(x)=\frac{1}{\sqrt{x}+1}$ satisfies the assumptions of the theorem is what allow us to do this exercise this way.

If you're function doesn't satisfy the hypothesis you can't use this technique. For example, if you considered $\displaystyle \int_{0}^{\infty}\sin (x) dx$, you couldn't do this problem using the reasoning we are now, because $\displaystyle \sin (x)$ goes below the x-axis and oscillates (so it is not strictly decreasing).

Enough with this technical stuff and back to the meat of the reasoning.

3. We know from 2 that $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x}+1}$ is convergent if and only if $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}+1}$. Equivalently, this means that if $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}+1}$ is divergent, then we will know that $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x}+1}$ is divergent. Our goal thus becomes proving $\displaystyle \sum_{n=0}^{\infty}\frac{1}{\sqrt{n}+1}$ is divergent.

4. Now $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ is divergent, because $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\geq \sum_{n=1}^{\infty}\frac{1}{n}$ and the series on the right of the inequality is the divergent "harmonic" series. The inequality comes from the fact that $\displaystyle \sqrt{n}\leq n$ for all $\displaystyle n$, which implies $\displaystyle \frac{1}{\sqrt{n}}\geq \frac{1}{n}$ for all $\displaystyle n$.

5. Using the Limit Comparison Test (see Limit comparison test - Wikipedia, the free encyclopedia) the fact that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ diverges (from step 4) implies that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+1}$ diverges. If you're unsure of what I mean here or exactly how to compute the necessary limit, let me know and I can explain it further.

We have just established that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+1}$ diverges, so (by step 3) we know $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x}+1}$ is divergent.

Notice that we didn't need to compute any antiderivatives. If this is the first time you're seeing something like this and are confused don't worry. It's perfectly normal to be confused by something you've never seen before; however, if you look it over enough, ask questions and challenge yourself it will make sense. I provided this alternative method to help illustrate that you can, on occasion, avoid tricky computations using nice theorems. Swing from the vines so you don't need to cut through the forrest!

Let me know if you have any questions about anything. Good luck!