# Splitting an integral for comparison test

I am looking to figure out if the following integral converges or diverges. $$\int_{0}^\infty{\frac{dx}{\sqrt{x}*(1+x)}}$$ = $$\int_{0}^\infty{\frac{dx}{\mathrm{x}^{1/2}+\mathrm{x}^{3/2}}}$$.

I have set it less than or equal to $$\int_{0}^\infty{\frac{dx}{\mathrm{x}^{3/2}}}$$. Since there is an asymptote at $$x = 0$$, I split this integral into $$\int_{0}^1{\frac{dx}{\mathrm{x}^{3/2}}}$$ + $$\int_{1}^\infty{\frac{dx}{\mathrm{x}^{3/2}}}$$.

However this results in a contradiction because the p test says the first part should diverge and the second part of this split integral should converge. Therefore, I cannot say anything about the convergence of the original integral. I'm not sure what to do at this point.

• If you split an integral and one part of it doesn't converge, the whole original integral diverges. Jan 20, 2019 at 22:16
• For future reference - if you want to put something with more than one character in an exponent or the like, use curly braces to enclose it. (This correction has already been edited into your post.) Also, \cdot is a nice multiplication dot, better looking than *. Jan 20, 2019 at 22:17
• @DonAntonio: True - but the comparison test returned results of "converges" for one part and "unknown" for the other; a one-sided comparison that says "less than something divergent" doesn't prove anything. What it means is that we need a better comparison for that part. Jan 20, 2019 at 22:19
• @jmerry I wrote about what the OP wrote in her(his) question: "the p test says the first part should diverge and the second part of this split integral should converge. Therefore, I cannot say anything about the convergence of the original integral" . This is inaccurate as if this was really the case the integral diverges. That's what I said Jan 20, 2019 at 22:21

$$\int\frac{dx}{\sqrt x(1+x)}=2\int\frac{d\sqrt x}{1+\left(\sqrt x\right)^2}=2\arctan\sqrt x+C\;\ldots\ldots$$
For $$1\le x<\infty$$, $$x^{\frac12}\le x^{\frac32}$$ and the denominator is dominated by the $$x^{\frac32}$$ term, so we compare to $$x^{-\frac32}$$. For $$0, $$x^{\frac12}\ge x^{\frac32}$$ and the denominator is dominated by the $$x^{\frac12}$$ term, so we compare to $$x^{-\frac12}$$ instead.