Convergence of Improper Integrals: Homework

[Mattofix]

Homework Statement

do the following integrals converge?

i) $$\int_0^{1}\frac{dx}{x^{3/2}e^{x}}$$

ii) $$\int_0^{1}\frac{x}{\sqrt{1-x^{2}}}dx$$

The Attempt at a Solution

looking at them i can guess that they both diverge - proving this is the hard part - this is what i have got but it doesn't prove anything...

i) $$\frac{1}{x^{3/2}}$$$$\geq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\geq{e^{-x}}$$

$$e^{-x}$$ converges
$$\frac{1}{e^{x}}$$$$\leq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\leq{x^{-3/2}$$

$$x^{-3/2}$$ divereges
ii) $$x\leq1$$

$$\frac{x}{\sqrt{1-x^{2}}}$$$$\leq{\frac{1}{\sqrt{1-x^{2}}}$$

$${\frac{1}{\sqrt{1-x^{2}}}$$ diverges (cannot prove it though)
$${\frac{1}{\sqrt{1-x^{2}}}\geq1$$

$${\frac{x}{\sqrt{1-x^{2}}}$$$$\geq x$$

x converges

any help would be much appreciated.

 

[e(ho0n3]

i) $$\frac{1}{x^{3/2}}$$$$\geq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\geq{e^{-x}}$$

$$e^{-x}$$ converges

I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

$$\frac{1}{e^{x}}$$$$\leq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\leq{x^{-3/2}$$

$$x^{-3/2}$$ divereges

Again, what does that tell you about the left-hand side? Ditto for ii).

 

[Mattofix]

I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

maybe I'm missing something but not much because the left hand side is greater than the right, the right converges, this does not mean the same for the left - it might diverge.

The same for all the others - the comparisions don't tell us much.

If all the inequalities were reversed then i wouldn’t have a problem.

i think...?

 

[e(ho0n3]

Yes. Exactly. You need to find an expression so that the inequalities are reversed.