2 (simple) Improper Integrals I'm stuck with

[maverick280857]

Hi everyone

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

1. <br /> \int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx<br />

2. <br /> \int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt<br />

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 &gt; 0 or p &lt; 2.

Similarly, I_{2} converges for p &gt; 1. Putting these together, the original integral converges for 1 &lt; p &lt; 2. Is this correct?

Second Problem

<br /> \int_{0}^{\infty} \frac{t^{p-1}}{1+t} dt = I_{1} + I_{2}<br />

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 &gt; 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 &lt; p &lt; 1.

 

[courtrigrad]

1. This is correctFor 2 how did you conclude that it I_{1} will converge for p &gt; 0? Use the limit comparison test.

 

[maverick280857]

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

 

[maverick280857]

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

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 &gt; 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!)

 

[maverick280857]

\frac{1}{t^{1-p}} converges when p &lt; 0

This is for 1 &lt; t &lt; \infty. What about the first integral?

 

[courtrigrad]

it converges for p&gt;0. so the whole integral diverges because p &lt; 0 and p &gt; 0 can't both be true.

 

[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.

 

[courtrigrad]

indeed, you are correct, it converges for 0&lt;p&lt;1. I used the same argument as in problem 1.

you have to set 1-p &lt; 1 for I_{1}, and 1-p &gt; 0 for I_{2}