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Improper integrals

Show that none of the following improper integrals exists.
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Shas72
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Original post by Shas72
Show that none of the following improper integrals exists.

15736471950967768615989860071610.jpgIam not able to understand this. Pls explain
Original post by Shas72
Iam not able to understand this. Pls explain


Do you understand what an improper integral is ?
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Shas72
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improper integrals are of two types. definite is with finite value and undefined has infinite value right?
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also should I just see the limits and determine whether its undefined or not
Original post by Shas72
improper integrals are of two types. definite is with finite value and undefined has infinite value right?


Not quite.

Definite integrals are those with limits on them.

Indefinite integrals are those without limits on them.

Improper integrals are definite integrals when at least one of the points over our range of integration is a bad point. For instance, infinities are bad because we don't know what happens there exactly. Also other points are those where the integrand is undefined, so if my integrand is 1/x and my range of integration is from -1 to 1, this is an issue because my integrand isn't defined at x=0 which is in the region of integration.

This motivates the need to instead look at what happens when we approach these bad points in the limit. So for instance,

∫1∞dxx\displaystyle \int_1^{\infty} \dfrac{dx}{\sqrt{x}} is replaced by

lim⁡b→∞∫1bdxx\displaystyle \lim_{b \to \infty} \int_1^b \dfrac{dx}{\sqrt{x}}

where we first work out the proper definite integral, and only then take the limit to see if it exists.

Similarly, if my range of integration was instead

∫01dxx\displaystyle \int_0^{1} \dfrac{dx}{\sqrt{x}}

then x=0 is a bad point and so we would replace it by

lim⁡a→0∫a1dxx\displaystyle \lim_{a \to 0} \int_a^{1} \dfrac{dx}{\sqrt{x}}

and likewise evaluate the proper definite integral before taking the limit.
(edited 4 years ago)
Original post by Shas72
Iam not able to understand this. Pls explain


So with the above post in mind, how can you rewrite

∫4∞6x.dx\displaystyle \int_4^{\infty} \dfrac{6}{\sqrt{x}}.dx

involving a limit of a proper, definite integral ?
(edited 4 years ago)
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Shas72
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thanks a lot!!! that was very very helpful
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image-84908363-27c7-44ed-8cd0-ba6e997d7ec21611810296173335850-compressed.jpg.jpeg
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is this right?
Original post by Shas72
...


Firstly, a minor note. I would avoid using XX and just use aa (for lower limit) and bb (for upper limit) just to be clearer. [of course, sometimes in a more algebraic example you might have a,b in the integrand as some constants in which case you should use different letters for the limits again]

Secondly, a major note. You are correct up to 12X−12412\sqrt{X} - 12\sqrt{4} but then you take the limit and it really doesn't make sense. If XX goes to positive infinity, then X\sqrt{X} does the same. Hence 12X→∞12\sqrt{X} \to \infty as a result. But you say it goes to zero instead for some reason ??

Also, even if it *did* go to zero, I'm not sure why you're saying "limit are positive" ?? What does that mean? Why must it be positive? It's not a correct justification for this question.
(edited 4 years ago)
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limits are positive as the lower limit is 0 and higher limit is infinity. if the limits were - infinity and 0 then the finite value would be -24.
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what would your method be to justify that this is undefined?
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Original post by RDKGames
Firstly, a minor note. I would avoid using XX and just use aa (for lower limit) and bb (for upper limit) just to be clearer. [of course, sometimes in a more algebraic example you might have a,b in the integrand as some constants in which case you should use different letters for the limits again]

Secondly, a major note. You are correct up to 12X−12412\sqrt{X} - 12\sqrt{4} but then you take the limit and it really doesn't make sense. If XX goes to positive infinity, then X\sqrt{X} does the same. Hence 12X→∞12\sqrt{X} \to \infty as a result. But you say it goes to zero instead for some reason ??

Also, even if it *did* go to zero, I'm not sure why you're saying "limit are positive" ?? What does that mean? Why must it be positive? It's not a correct justification for this question.

So should I write if X goes to 0 , 12 square root X goes to infinity
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Original post by RDKGames
Firstly, a minor note. I would avoid using XX and just use aa (for lower limit) and bb (for upper limit) just to be clearer. [of course, sometimes in a more algebraic example you might have a,b in the integrand as some constants in which case you should use different letters for the limits again]

Secondly, a major note. You are correct up to 12X−12412\sqrt{X} - 12\sqrt{4} but then you take the limit and it really doesn't make sense. If XX goes to positive infinity, then X\sqrt{X} does the same. Hence 12X→∞12\sqrt{X} \to \infty as a result. But you say it goes to zero instead for some reason ??

Also, even if it *did* go to zero, I'm not sure why you're saying "limit are positive" ?? What does that mean? Why must it be positive? It's not a correct justification for this question.

Ok so I got it. If X tends to infinity 12 square root X also tends to infinity. Hence it is undefined
Original post by Shas72
Ok so I got it. If X tends to infinity 12 square root X also tends to infinity. Hence it is undefined


Exactly this. :smile:

If X→∞X \to \infty then 12X−12412\sqrt{X} - 12\sqrt{4} tends to infinity, hence this integral does not exist.
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ok I understood. thanks a lottttttt

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