# Homework Help: 2 (simple) Improper Integrals I'm stuck with

1. Oct 16, 2006

### maverick280857

Hi everyone

Here are two improper integrals and I have to find the values of $p$ for which they are convergent

1. $$\int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx$$

2. $$\int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt$$

As I do not know the answers and am somewhat stuck with the reasoning, I would be grateful if someone could check these for me and let me know the correct answers

First Problem

$$\int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx = I_{1} + I_{2}$$

where

$$I_{1} = \int_{0}^{1} \frac{1-e^{-x}}{x^{p}}dx$$
$$I_{2} = \int_{1}^{\infty} \frac{1-e^{-x}}{x^{p}}dx$$

Now,

$$\frac{1-e^{-x}}{x^{p}} \leq \frac{x}{x^{p}} = \frac{1}{x^{p-1}}$$

So by comparison test (or limit comparison test) $I_{1}$ converges for $2-p > 0$ or $p < 2$.

Similarly, $I_{2}$ converges for $p > 1$. Putting these together, the original integral converges for $1 < p < 2$. Is this correct?

Second Problem

$$\int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt = I_{1} + I_{2}$$

where

$$I_{1} = \int_{0}^{1} \frac{t^{p-1}}{1+t} dt$$
$$I_{2} = \int_{1}^{\infty} \frac{t^{p-1}}{1+t} dt$$

Since,

$$\frac{t^{p-1}}{1+t} \leq t^{p-1}$$

by comparison test, we can see that $I_{1}$ will converge for $p > 0$. But this reasoning applied to $I_{2}$ yields a wrong answer. Whats going wrong here?

I think the correct answer to second problem is $0 < p < 1$.

2. Oct 16, 2006

1. This is correct

For 2 how did you conclude that it $$I_{1}$$ will converge for $$p > 0$$? Use the limit comparison test.

Last edited: Oct 16, 2006
3. Oct 16, 2006

### maverick280857

Thanks ...please...could you also have a look at the second one.

4. Oct 16, 2006

### maverick280857

My question then is two-fold.

First if I compare with $t^{p-1}$ then

$$\lim_{t \rightarrow 0}}\left(\frac{t^{p-1}}{1+t}\right)\left(\frac{1}{t^{p-1}}\right) = 1$$

So the integral converges if $\int_{0}^{1}t^{p-1}dt$ converges, i.e. for $p > 0$. (I think something's going wrong here...can you suggest a more general method? I think that by comparing with a power of p, I am forcing only that solution for which the right hand integral is convergent, which depends on p!)

5. Oct 16, 2006

### maverick280857

This is for $1 < t < \infty$. What about the first integral?

6. Oct 16, 2006

it converges for $$p>0$$. so the whole integral diverges because $$p < 0$$ and $$p > 0$$ cant both be true.

Last edited: Oct 16, 2006
7. Oct 16, 2006

### maverick280857

No, it actually converges:

Suppose I take p = 0.5, then the integral is simply,

$$\int_{0}^{\infty} \frac{1}{\sqrt{x}(1+x)}dx = \int_{0}^{\infty} \frac{2}{(1+t^{2})}dt$$

which converges to $\pi$.

Last edited: Oct 16, 2006
8. Oct 16, 2006

indeed, you are correct, it converges for $$0<p<1$$. I used the same argument as in problem 1.
you have to set $$1-p < 1$$ for $$I_{1}$$, and $$1-p > 0$$ for $$I_{2}$$