Integral Convergence: Examining 1/(e^x sqrt(x))

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Homework Help Overview

The discussion revolves around the convergence or divergence of the integral \(\int_1^\infty \frac{1}{e^x \sqrt{x}} \, dx\). Participants are examining the conditions under which this integral can be evaluated without directly solving it, focusing on comparison tests and the behavior of the integrand.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of comparison tests, noting that \(\frac{1}{\sqrt{x} e^x}\) is less than \(\frac{1}{e^x}\), which is known to converge. There are questions about the original poster's reasoning and presentation style, as well as the validity of their integral transcription.

Discussion Status

There is a consensus among some participants that the original poster's reasoning appears sound, yet they are puzzled by the feedback received on their exam. Suggestions for improving clarity in presentation have been offered, and there is an ongoing exploration of potential misunderstandings regarding the integral's setup.

Contextual Notes

Participants note that the original poster did not receive specific feedback on why their answer was marked incorrect, leading to further inquiry into the correctness of their approach and the clarity of their explanation.

Jeanclaud
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http://‪C:\Users\johny\Downloads\q4.jpg 1. Homework Statement
Hi, so the question is I have to tell if this integral diverges or converges.(without solving it)
integral(1/(e^x sqrt(x)))dx from 1 to +inf

Homework Equations


integration techniques.

The Attempt at a Solution


my answer: let 1/e^x >1/(e^x sqrt(x))
then I solved the definite integral(1/e^x)from 1 to +inf and got 1/e which means it converges.
so the given integral has to converge also since it is smaller than the 1/e^x.
that was my answer in the exam but they considered it all wrong so please can anybody tell me the reason. Thank you.
 
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Looking at the form of this integral $$ \int_1^\infty \frac 1 { \sqrt x e^x} \, dx $$

it seems fairly clear we're expecting convergence here - so for the comparison test we are looking for a larger integral on the domain that converges. We have
$$ \frac 1 { \sqrt x e^x } \leq \frac 1 { e^x } $$ on this domain. Everything's all positive so we're good to go.

The integral $$ \int_1^\infty \frac 1 { e^x } \, dx $$ is clearly convergent, so our integral of interest is also convergent.

I can't see why your answer was considered to be incorrect either :sorry:
 
Jeanclaud said:
so the given integral has to converge also since it is smaller than the 1/e^x.
that was my answer in the exam but they considered it all wrong
Was there any reason given why your answer was "all wrong"?
 
Mark44 said:
Was there any reason given why your answer was "all wrong"?
nope.
 

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Well, I can't see why your answer was considered to be wrong. Just a silly thought; you did actually copy the integral down from the exam paper correctly?

That's the only thing I can think of because your answer looks OK to me.

You could probably structure your answer a bit better - does your answer look anything like the worked examples in your textbooks in terms of how it's laid out? Try to copy this 'style' and set out your answers in a clear step-by-step fashion and explain what you're doing (only takes a few words here and there). It's a bit of a pain to do this initially but it becomes easier with practice and eventually becomes second nature - and you (should) find your understanding and maybe even your marks improve the more you do this :woot:
 
Jeanclaud said:
nope.

I second the remarks of Simon Phoenix regarding your presentation style. I would add that some of the things you wrote are technically wrong without further qualification. In particular, the inequality ##e^{-x} > e^{-x}/\sqrt{x}## is false when ##x < 1##, but true when ##x > 1##. You could say something like "since we want x > 1, ..." and then what you wrote would be correct. Just a few words of explanation is all you need; it would take < 5 seconds to write them.
 
thanks you.
 
Ray Vickson said:
second the remarks of Simon Phoenix regarding your presentation style. I would add that some of the things you wrote are technically wrong without further qualification. In particular, the inequality e−x > e−x/ √x is false when x < 1, but true when x > 1.
But since the interval of integration is ##[1, \infty)##, I don't think it's absolutely necessary to stipulate that x > 1.
 
Last edited:
  • #10
Mark44 said:
But since the interval of integration is ##[1, \infty)##, I don't think it's absolutely necessary to stipulate that x > 1.

We are all just trying to figure out why he lost marks on correct work.
 

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