Comparing integral resolutions using Wolfram Alpha / Mathematica

Question

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.

Answer 1

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 $\frac{-2}{\sqrt{x}}$.

Answer 2

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 $g'(x)=x^{-3/2}=\frac{1}{\sqrt{x}^3}=\frac{1}{x\sqrt{x}}$.