# Improper Integrals

1. Mar 3, 2008

### Mattofix

1. The problem statement, all variables and given/known data

do the following integrals converge?

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

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

3. 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.

Last edited: Mar 3, 2008
2. Mar 3, 2008

### e(ho0n3

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?

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

3. Mar 3, 2008

### Mattofix

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...?

Last edited: Mar 3, 2008
4. Mar 3, 2008

### e(ho0n3

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