Equations are in $\LaTeX$ format; I'm still trying to understand how MathJax works.

Given the following integral:

$\int_0^{+\infty } \frac{1}{x \sqrt{x}} \, dx$

I'm pretty sure that does not converge. And if the integral was a indefinite one, also I'm almost sure the result is $\ln(\sqrt x)$ using substitution. If I am correct, why wolfram alpha says that the result is $\frac{-2}{\sqrt x}$ instead of $\ln(\sqrt x)$?

MathWay shows the correct results for both definite and indefinite integrals. I'm new to any mathematical soft like Wolfram Alpha/Mathematica, and my idea is to test my pen and paper results with software that checks my results.

Any hints will be greatly appreciated. Thanks.

  • $\begingroup$ The integral does converge (if you start it from 1 instead of 0), and WolframAlpha is correct. Remember that x sqrt{x} = x^{3/2}. $\endgroup$ – Qiaochu Yuan Nov 26 '10 at 22:56
  • $\begingroup$ Interesting If WolframAlpha is right, Mathway is wrong: try this ∫[1/(x*√(x)),x] on Mathway and see the result. Anyway, You are absolutely right. Now I see it x^1 * x^(1/2) = x^(3/2) $\endgroup$ – Markust Nov 26 '10 at 23:05
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    $\begingroup$ "MathWay" claimed the answer was ln(|x|) when I tried it. I do not think it is worth paying attention to. Whatever else you can say about Wolfram Alpha, it is generally very good at integrals. $\endgroup$ – Qiaochu Yuan Nov 26 '10 at 23:07

The indefinite integral is $$\int \frac{1}{x\sqrt{x}} \, dx = \int \frac{1}{x^{^{3/2}}}\, dx = \frac{-2}{\sqrt{x}} + C$$ I suppose you made a mistake when integrating it by making a wrong substitution $u = \sqrt{x}$. Probably that's how you got the $\ln{\sqrt{x}}$.

About the convergence issue, you're right, the integral is divergent as you can see by taking the corresponding limit at $0$ of the indefinite integral $\frac{-2}{\sqrt{x}}$.

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    $\begingroup$ Making the substitution $u=\sqrt x$ is not "wrong", it is just inefficient. It leads to $2\int \frac{1}{u^2}du$. $\endgroup$ – Jonas Meyer Nov 26 '10 at 22:59
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    $\begingroup$ @Jonas I was just making the point that probably that's how the OP arrived at $\ln{\sqrt{x}}$. $\endgroup$ – Adrián Barquero Nov 26 '10 at 23:05
  • $\begingroup$ Yep, I just did it again, thanks $\endgroup$ – Markust Nov 26 '10 at 23:06

If you are at the point of computing integrals, then you are probably already comfortable with computing derivatives. The way to check whether a potential antiderivative is correct is to take its derivative. If you take the derivative of $f(x)=\ln(\sqrt x)$ (either by first simplifying to $\frac{1}{2}\ln(x)$ or using the chain rule) you get $f'(x)=\frac{1}{2x}$. If you take the derivative of $g(x)=-\frac{2}{\sqrt x}=-2x^{-1/2}$, you get $g'(x)=x^{-3/2}=\frac{1}{\sqrt{x}^3}=\frac{1}{x\sqrt{x}}$.


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