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

1. [tex]

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

[/tex]

2. [tex]

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

[/tex]

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

[tex]\int_{0}^{\infty} \frac{1-e^{-x}}{x^{p}}dx = I_{1} + I_{2}[/tex]

where

[tex]I_{1} = \int_{0}^{1} \frac{1-e^{-x}}{x^{p}}dx[/tex]

[tex]I_{2} = \int_{1}^{\infty} \frac{1-e^{-x}}{x^{p}}dx[/tex]

Now,

[tex]\frac{1-e^{-x}}{x^{p}} \leq \frac{x}{x^{p}} = \frac{1}{x^{p-1}}[/tex]

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

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

Second Problem

[tex]

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

[/tex]

where

[tex]I_{1} = \int_{0}^{1} \frac{t^{p-1}}{1+t} dt[/tex]

[tex]I_{2} = \int_{1}^{\infty} \frac{t^{p-1}}{1+t} dt[/tex]

Since,

[tex]\frac{t^{p-1}}{1+t} \leq t^{p-1}[/tex]

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

I think the correct answer to second problem is [itex] 0 < p < 1[/itex].

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

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