Improper integrals (1 Viewer)

leehuan

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The answer is obviously that it does not, but I want to check if my proof is flawed.







I have a feeling I may have needed to consider the negative infinity portion of this integral?
 
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InteGrand

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The answer is obviously that it does not, but I want to check if my proof is flawed.







I have a feeling I may have needed to consider the negative infinity portion of this integral?
The Cauchy principal value would be 0. Otherwise it's an undefined expression -inf + inf.

Don't need to use any comparison tests, we can find an antiderivative using inspection or a u-substitution.
 
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Paradoxica

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The answer is obviously that it does not, but I want to check if my proof is flawed.







I have a feeling I may have needed to consider the negative infinity portion of this integral?
Does it make sense to ask convergence of an integral whose CPV is 0?

In any case, you can reduce the two cases to one by using the parity of the function to your advantage.
 

leehuan

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They didn't teach us about the Cauchy principal value. I only found that on WolframAlpha.

Edit: Though they did make us evaluate this
 
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leehuan

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Bit of guidance needed here.

 
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leehuan

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(I know this is the Gamma function)

I had an idea but then I got lost. My last step was rewriting the integral as

 
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InteGrand

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(I know this is the Gamma function)

I had an idea but then I got lost. My last step was rewriting the integral as

The -t in the exponential should become -1/x , rather than 1/x.
 
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Drongoski

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In transformed integral: ??


ps
InteGrand beaten me to it.
 

seanieg89

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So you have the integrand and we want to show it is integrable on the non-negative reals.

The behaviour near zero is not a problem at all, since as (essentially because exponentials dominate polynomials, there are many ways you could justify this). So it suffices to check that g be integrable on .

This follows by comparison to . (In fact we can use this argument to show that the integral is convergent if and ONLY if ).
 

leehuan

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Having already shown



The method the answers provided is not immediately obvious to me and I don't understand the lateral thinking required. Can someone please explain the intuition required to see this method or just provide an easier pathway?



 

InteGrand

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Having already shown



The method the answers provided is not immediately obvious to me and I don't understand the lateral thinking required. Can someone please explain the intuition required to see this method or just provide an easier pathway?



Yeah it's essentially a comparison test.

For all t large enough, the integrand in the second integral is greater than the first (since t^2 + t > 2t for all t large enough, say all t > 1).

Since the former integral taken from 1 to oo diverges to +oo (from previous part of Q.), by the comparison test, so does the latter one when taken from 1 to oo.

And remembering the finite lower limit here is irrelevant since the integrand is continuous, we're done.
 

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