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

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SUMMARY

The integral ∫(1/(e^x sqrt(x)))dx from 1 to +∞ converges based on the comparison test with the integral ∫(1/e^x)dx, which is known to converge. The discussion highlights that the original poster's reasoning was valid, as 1/(e^x sqrt(x)) is less than 1/e^x for x > 1. However, the original poster's answer was marked incorrect, possibly due to presentation issues or minor technical inaccuracies in their explanation.

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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.
 
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  • #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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