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How would I approach solving this improper integral using a Riemann integral? [closed]

0

1

$begingroup$

$$int_1^infty frac1x^2mathrm dx$$

We're asked what happens if we use the typical Riemann integral as defined on this improper integral. We can't use typical Riemann integrals to solve improper integrals, so to make it finite we would use a dummy variable as $y$ goes to infinity, then solve it as $-frac1y+1=1$, but what else is this question asking in using the Riemann definition?

calculus

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edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

$endgroup$

closed as unclear what you're asking by José Carlos Santos, RRL, Lord Shark the Unknown, mrtaurho, Javi Mar 25 at 11:27

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

0

1

$begingroup$

$$int_1^infty frac1x^2mathrm dx$$

We're asked what happens if we use the typical Riemann integral as defined on this improper integral. We can't use typical Riemann integrals to solve improper integrals, so to make it finite we would use a dummy variable as $y$ goes to infinity, then solve it as $-frac1y+1=1$, but what else is this question asking in using the Riemann definition?

calculus

share|cite|improve this question

edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

$endgroup$

closed as unclear what you're asking by José Carlos Santos, RRL, Lord Shark the Unknown, mrtaurho, Javi Mar 25 at 11:27

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

$$int_1^infty frac1x^2mathrm dx$$

We're asked what happens if we use the typical Riemann integral as defined on this improper integral. We can't use typical Riemann integrals to solve improper integrals, so to make it finite we would use a dummy variable as $y$ goes to infinity, then solve it as $-frac1y+1=1$, but what else is this question asking in using the Riemann definition?

calculus

share|cite|improve this question

edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

$endgroup$

share|cite|improve this question

edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

share|cite|improve this question

edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

edited Mar 25 at 8:41

mrtaurho

6,15271641

edited Mar 25 at 8:41

mrtaurho

6,15271641

asked Mar 24 at 18:33

happysainthappysaint

112

asked Mar 24 at 18:33

happysainthappysaint

112

1 Answer
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$begingroup$

Since the Riemann integral is only defined for closed and bounded intervals, it does not make sense to use the Riemann integral here. Indeed, as you noted, one has to make use of the "improper" Riemann integral which involves taking a limit. More precisely, the symbol $int_1^infty frac1x^2,mathrmdx$ should be interpreted as the limit
beginalign*
int_1^infty frac1x^2,mathrmdx = lim_y to inftyint_1^yfrac1x^2,mathrmdx
&= lim_y to infty left[ left.-frac1xrightvert_x=1^x=y right]\
&= lim_y to infty left( 1 - frac1yright)\
&= 1.
endalign*
The idea here is partially extend the Riemann integral (which, again, is only defined on bounded intervals) to "nice functions" on possibly unbounded intervals by taking a limit. Thus, an improper integral is really just a limit of Riemann integrals. This is likely what is meant by "using the Riemann integral".

As for what the question means by "what happens if we use the typical Riemann definition", my guess is that the purpose of that question is to point out that the definition of the Riemann integral only applies with bounded intervals.

share|cite|improve this answer

edited Mar 24 at 19:09

answered Mar 24 at 19:04

rolandcyprolandcyp

1

$endgroup$

add a comment | 

1

$begingroup$

Since the Riemann integral is only defined for closed and bounded intervals, it does not make sense to use the Riemann integral here. Indeed, as you noted, one has to make use of the "improper" Riemann integral which involves taking a limit. More precisely, the symbol $int_1^infty frac1x^2,mathrmdx$ should be interpreted as the limit
beginalign*
int_1^infty frac1x^2,mathrmdx = lim_y to inftyint_1^yfrac1x^2,mathrmdx
&= lim_y to infty left[ left.-frac1xrightvert_x=1^x=y right]\
&= lim_y to infty left( 1 - frac1yright)\
&= 1.
endalign*
The idea here is partially extend the Riemann integral (which, again, is only defined on bounded intervals) to "nice functions" on possibly unbounded intervals by taking a limit. Thus, an improper integral is really just a limit of Riemann integrals. This is likely what is meant by "using the Riemann integral".

As for what the question means by "what happens if we use the typical Riemann definition", my guess is that the purpose of that question is to point out that the definition of the Riemann integral only applies with bounded intervals.

share|cite|improve this answer

edited Mar 24 at 19:09

answered Mar 24 at 19:04

rolandcyprolandcyp

1

$endgroup$

add a comment | 

1

$begingroup$

Since the Riemann integral is only defined for closed and bounded intervals, it does not make sense to use the Riemann integral here. Indeed, as you noted, one has to make use of the "improper" Riemann integral which involves taking a limit. More precisely, the symbol $int_1^infty frac1x^2,mathrmdx$ should be interpreted as the limit
beginalign*
int_1^infty frac1x^2,mathrmdx = lim_y to inftyint_1^yfrac1x^2,mathrmdx
&= lim_y to infty left[ left.-frac1xrightvert_x=1^x=y right]\
&= lim_y to infty left( 1 - frac1yright)\
&= 1.
endalign*
The idea here is partially extend the Riemann integral (which, again, is only defined on bounded intervals) to "nice functions" on possibly unbounded intervals by taking a limit. Thus, an improper integral is really just a limit of Riemann integrals. This is likely what is meant by "using the Riemann integral".

As for what the question means by "what happens if we use the typical Riemann definition", my guess is that the purpose of that question is to point out that the definition of the Riemann integral only applies with bounded intervals.

share|cite|improve this answer

edited Mar 24 at 19:09

answered Mar 24 at 19:04

rolandcyprolandcyp

1

$endgroup$

add a comment | 

$begingroup$

Since the Riemann integral is only defined for closed and bounded intervals, it does not make sense to use the Riemann integral here. Indeed, as you noted, one has to make use of the "improper" Riemann integral which involves taking a limit. More precisely, the symbol $int_1^infty frac1x^2,mathrmdx$ should be interpreted as the limit
beginalign*
int_1^infty frac1x^2,mathrmdx = lim_y to inftyint_1^yfrac1x^2,mathrmdx
&= lim_y to infty left[ left.-frac1xrightvert_x=1^x=y right]\
&= lim_y to infty left( 1 - frac1yright)\
&= 1.
endalign*
The idea here is partially extend the Riemann integral (which, again, is only defined on bounded intervals) to "nice functions" on possibly unbounded intervals by taking a limit. Thus, an improper integral is really just a limit of Riemann integrals. This is likely what is meant by "using the Riemann integral".

As for what the question means by "what happens if we use the typical Riemann definition", my guess is that the purpose of that question is to point out that the definition of the Riemann integral only applies with bounded intervals.

share|cite|improve this answer

edited Mar 24 at 19:09

answered Mar 24 at 19:04

rolandcyprolandcyp

1

$endgroup$

share|cite|improve this answer

edited Mar 24 at 19:09

answered Mar 24 at 19:04

rolandcyprolandcyp

1

answered Mar 24 at 19:04

rolandcyprolandcyp

1

answered Mar 24 at 19:04

rolandcyprolandcyp

1