Mixing
How do integrals relate to converging and diverging series?
$begingroup$
When $n$ is an integer greater than $1$,
$int_0^1{frac{1}{x^n}}dx$
The answer key says the answer is $infty$.
Now I have experience with integration, finding convergence/divergence of series, etc. But, I'm not seeing the connection with series. $frac{1}{x^n}$ is clearly the form of the power series, but how do I connect it with this integral?
This leads to a more general question about the relationship between integrals and series, which I have searched for. I found some decent answers here, but I am certainly open to any thoughts about that topic.
Anyways, let me know what you think about solving that integral at the top.
calculus integration sequences-and-series
edited Jan 15 at 19:16
David G. Stork
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
$endgroup$
add a comment |
$begingroup$
When $n$ is an integer greater than $1$,
$int_0^1{frac{1}{x^n}}dx$
The answer key says the answer is $infty$.
Now I have experience with integration, finding convergence/divergence of series, etc. But, I'm not seeing the connection with series. $frac{1}{x^n}$ is clearly the form of the power series, but how do I connect it with this integral?
This leads to a more general question about the relationship between integrals and series, which I have searched for. I found some decent answers here, but I am certainly open to any thoughts about that topic.
Anyways, let me know what you think about solving that integral at the top.
calculus integration sequences-and-series
edited Jan 15 at 19:16
David G. Stork
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
$endgroup$
$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28
add a comment |
$begingroup$
When $n$ is an integer greater than $1$,
$int_0^1{frac{1}{x^n}}dx$
The answer key says the answer is $infty$.
Now I have experience with integration, finding convergence/divergence of series, etc. But, I'm not seeing the connection with series. $frac{1}{x^n}$ is clearly the form of the power series, but how do I connect it with this integral?
This leads to a more general question about the relationship between integrals and series, which I have searched for. I found some decent answers here, but I am certainly open to any thoughts about that topic.
Anyways, let me know what you think about solving that integral at the top.
calculus integration sequences-and-series
edited Jan 15 at 19:16
David G. Stork
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
$endgroup$
When $n$ is an integer greater than $1$,
$int_0^1{frac{1}{x^n}}dx$
The answer key says the answer is $infty$.
Now I have experience with integration, finding convergence/divergence of series, etc. But, I'm not seeing the connection with series. $frac{1}{x^n}$ is clearly the form of the power series, but how do I connect it with this integral?
This leads to a more general question about the relationship between integrals and series, which I have searched for. I found some decent answers here, but I am certainly open to any thoughts about that topic.
Anyways, let me know what you think about solving that integral at the top.
calculus integration sequences-and-series
calculus integration sequences-and-series
edited Jan 15 at 19:16
David G. Stork
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
edited Jan 15 at 19:16
David G. Stork
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
edited Jan 15 at 19:16
David G. Stork
11k41432
edited Jan 15 at 19:16
David G. Stork
11k41432
edited Jan 15 at 19:16
David G. Stork
11k41432
11k41432
asked Jan 15 at 19:09
AddisonAddison
314
asked Jan 15 at 19:09
AddisonAddison
314
asked Jan 15 at 19:09
AddisonAddison
314
314
$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28
add a comment |
$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28
$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$int_0^1{frac{1}{x^n}}dx$ is an improper integral, because for $x to 0$ we have $frac{1}{x^n} to infty$.
So, by definition, the value of the integral for $n>1$ is given by:
$$
int_0^1{frac{1}{x^n}}dx=
$$
$$
=
frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}=frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]
$$
that, with the substitution $y=1/x$ becomes:
$$
=frac{1}{n-1}left[lim_{y to infty};(y^{n-1})-1 right]
$$
Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.
The last step comes from:
$$
lim_{x to 0}(x^{1-n})
$$
$$
=lim_{x to 0}frac{1}{x^{n-1}}
$$
$$
=lim_{x to 0}left(frac{1}{x}right)^{n-1}=
$$
and the substitution:
$$
frac{1}{x}=y
$$
for which we have that if $ xto 0$ than $y to infty$.
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
$endgroup$
– Addison
Jan 15 at 20:55
|
show 3 more comments
$begingroup$
let $y = x^{-1}$
$dy = -x^{-2} dx\
dx = - y^{-2} dy$
$int_{0}^1 x^{-n} dx = int_1^{infty} y^{n-2} dx $
If $n ge 2$ then the integral is clearly divergent as
$lim_limits {yto infty} y^{n-2}ne 0$
If $n < 2$ then $y^{n-2}$ is montonic and decreasing.
$sum_limits {k=1}^{infty} (k^{n-2}) leint_1^{infty} y^{n-2} dyle sum_limits {k=2}^{infty} (k^{n-2})$
if $nge1$
$sum_limits {k=1}^{infty} (k^{n-2})$ diverges
if $n < 1$
$sum_limits {k=2}^{infty} (k^{n-2})$ converges
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$int_0^1{frac{1}{x^n}}dx$ is an improper integral, because for $x to 0$ we have $frac{1}{x^n} to infty$.
So, by definition, the value of the integral for $n>1$ is given by:
$$
int_0^1{frac{1}{x^n}}dx=
$$
$$
=
frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}=frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]
$$
that, with the substitution $y=1/x$ becomes:
$$
=frac{1}{n-1}left[lim_{y to infty};(y^{n-1})-1 right]
$$
Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.
The last step comes from:
$$
lim_{x to 0}(x^{1-n})
$$
$$
=lim_{x to 0}frac{1}{x^{n-1}}
$$
$$
=lim_{x to 0}left(frac{1}{x}right)^{n-1}=
$$
and the substitution:
$$
frac{1}{x}=y
$$
for which we have that if $ xto 0$ than $y to infty$.
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
$endgroup$
– Addison
Jan 15 at 20:55
|
show 3 more comments
$begingroup$
$int_0^1{frac{1}{x^n}}dx$ is an improper integral, because for $x to 0$ we have $frac{1}{x^n} to infty$.
So, by definition, the value of the integral for $n>1$ is given by:
$$
int_0^1{frac{1}{x^n}}dx=
$$
$$
=
frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}=frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]
$$
that, with the substitution $y=1/x$ becomes:
$$
=frac{1}{n-1}left[lim_{y to infty};(y^{n-1})-1 right]
$$
Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.
The last step comes from:
$$
lim_{x to 0}(x^{1-n})
$$
$$
=lim_{x to 0}frac{1}{x^{n-1}}
$$
$$
=lim_{x to 0}left(frac{1}{x}right)^{n-1}=
$$
and the substitution:
$$
frac{1}{x}=y
$$
for which we have that if $ xto 0$ than $y to infty$.
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
$endgroup$
– Addison
Jan 15 at 20:55
|
show 3 more comments
$begingroup$
$int_0^1{frac{1}{x^n}}dx$ is an improper integral, because for $x to 0$ we have $frac{1}{x^n} to infty$.
So, by definition, the value of the integral for $n>1$ is given by:
$$
int_0^1{frac{1}{x^n}}dx=
$$
$$
=
frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}=frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]
$$
that, with the substitution $y=1/x$ becomes:
$$
=frac{1}{n-1}left[lim_{y to infty};(y^{n-1})-1 right]
$$
Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.
The last step comes from:
$$
lim_{x to 0}(x^{1-n})
$$
$$
=lim_{x to 0}frac{1}{x^{n-1}}
$$
$$
=lim_{x to 0}left(frac{1}{x}right)^{n-1}=
$$
and the substitution:
$$
frac{1}{x}=y
$$
for which we have that if $ xto 0$ than $y to infty$.
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
$endgroup$
$int_0^1{frac{1}{x^n}}dx$ is an improper integral, because for $x to 0$ we have $frac{1}{x^n} to infty$.
So, by definition, the value of the integral for $n>1$ is given by:
$$
int_0^1{frac{1}{x^n}}dx=
$$
$$
=
frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}=frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]
$$
that, with the substitution $y=1/x$ becomes:
$$
=frac{1}{n-1}left[lim_{y to infty};(y^{n-1})-1 right]
$$
Note that here we have a limit that is the limit of a sequence, not the sum of a series, and, for $n>1$, this sequence is clearly divergent.
The last step comes from:
$$
lim_{x to 0}(x^{1-n})
$$
$$
=lim_{x to 0}frac{1}{x^{n-1}}
$$
$$
=lim_{x to 0}left(frac{1}{x}right)^{n-1}=
$$
and the substitution:
$$
frac{1}{x}=y
$$
for which we have that if $ xto 0$ than $y to infty$.
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
edited Jan 15 at 20:43
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
answered Jan 15 at 19:56
Emilio NovatiEmilio Novati
52k43474
52k43474
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
$endgroup$
– Addison
Jan 15 at 20:55
|
show 3 more comments
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
$endgroup$
– Addison
Jan 15 at 20:55
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
So I see how in the first step you simply integrated and plugged in the limits. I see you set the second term, as a limit, and that's fair enough. However, I don't see how you were able to get rid of the first term as you went to the next step. Also why does the denominator turn from 1-n to n-1?
$endgroup$
– Addison
Jan 15 at 20:16
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
I added to my answer. I hope it's useful :)
$endgroup$
– Emilio Novati
Jan 15 at 20:28
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
sorry, I'm still having trouble getting this. So I see that you integrated and put it into two terms: $frac{1}{1-n}(1)^{1-n}-lim_{x to 0}frac{1}{1-n}x^{1-n}$. I'm guessing the reason the denominator of $lim_{x to 0}frac{1}{1-n}x^{1-n}$ becomes n-1 is because of the negative in front of the limit, making things negative. But then how did you get the -1 in $frac{1}{n-1}left[lim_{x to 0};(x^{1-n})-1 right]$. The term $frac{1}{1-n}(1)^{1-n}$ doesn't simplify to -1.
$endgroup$
– Addison
Jan 15 at 20:47
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
$begingroup$
Note: $-frac{1}{1-n}=frac{1}{n-1}$ and $frac{1}{n-1}$ is a common factor.
$endgroup$
– Emilio Novati
Jan 15 at 20:53
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So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
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– Addison
Jan 15 at 20:55
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So what happens to the exponent of 1-n ? Doesn't that make the sign change? How can we factor out $frac{1}{n-1}$ if 1 is alternating?
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– Addison
Jan 15 at 20:55
|
show 3 more comments
$begingroup$
let $y = x^{-1}$
$dy = -x^{-2} dx\
dx = - y^{-2} dy$
$int_{0}^1 x^{-n} dx = int_1^{infty} y^{n-2} dx $
If $n ge 2$ then the integral is clearly divergent as
$lim_limits {yto infty} y^{n-2}ne 0$
If $n < 2$ then $y^{n-2}$ is montonic and decreasing.
$sum_limits {k=1}^{infty} (k^{n-2}) leint_1^{infty} y^{n-2} dyle sum_limits {k=2}^{infty} (k^{n-2})$
if $nge1$
$sum_limits {k=1}^{infty} (k^{n-2})$ diverges
if $n < 1$
$sum_limits {k=2}^{infty} (k^{n-2})$ converges
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
add a comment |
$begingroup$
let $y = x^{-1}$
$dy = -x^{-2} dx\
dx = - y^{-2} dy$
$int_{0}^1 x^{-n} dx = int_1^{infty} y^{n-2} dx $
If $n ge 2$ then the integral is clearly divergent as
$lim_limits {yto infty} y^{n-2}ne 0$
If $n < 2$ then $y^{n-2}$ is montonic and decreasing.
$sum_limits {k=1}^{infty} (k^{n-2}) leint_1^{infty} y^{n-2} dyle sum_limits {k=2}^{infty} (k^{n-2})$
if $nge1$
$sum_limits {k=1}^{infty} (k^{n-2})$ diverges
if $n < 1$
$sum_limits {k=2}^{infty} (k^{n-2})$ converges
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
$endgroup$
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
add a comment |
$begingroup$
let $y = x^{-1}$
$dy = -x^{-2} dx\
dx = - y^{-2} dy$
$int_{0}^1 x^{-n} dx = int_1^{infty} y^{n-2} dx $
If $n ge 2$ then the integral is clearly divergent as
$lim_limits {yto infty} y^{n-2}ne 0$
If $n < 2$ then $y^{n-2}$ is montonic and decreasing.
$sum_limits {k=1}^{infty} (k^{n-2}) leint_1^{infty} y^{n-2} dyle sum_limits {k=2}^{infty} (k^{n-2})$
if $nge1$
$sum_limits {k=1}^{infty} (k^{n-2})$ diverges
if $n < 1$
$sum_limits {k=2}^{infty} (k^{n-2})$ converges
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
$endgroup$
let $y = x^{-1}$
$dy = -x^{-2} dx\
dx = - y^{-2} dy$
$int_{0}^1 x^{-n} dx = int_1^{infty} y^{n-2} dx $
If $n ge 2$ then the integral is clearly divergent as
$lim_limits {yto infty} y^{n-2}ne 0$
If $n < 2$ then $y^{n-2}$ is montonic and decreasing.
$sum_limits {k=1}^{infty} (k^{n-2}) leint_1^{infty} y^{n-2} dyle sum_limits {k=2}^{infty} (k^{n-2})$
if $nge1$
$sum_limits {k=1}^{infty} (k^{n-2})$ diverges
if $n < 1$
$sum_limits {k=2}^{infty} (k^{n-2})$ converges
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
answered Jan 15 at 19:26
Doug MDoug M
45.2k31854
45.2k31854
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
add a comment |
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
I don't see how you were able to do this step: $int_{0}^1 x^{-n} $dx = $int_1^{infty} y^{n-2} dx$
$endgroup$
– Addison
Jan 15 at 19:30
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
$begingroup$
The limits of integration, $lim_limits {xto 0^+} y(x) = infty, y(1) = 1.$ Plugging all of these we get $int_{infty}^1 (y^n)(-y^{-2}) dy. $ We can flip the limits of integration if we also flip the sign.
$endgroup$
– Doug M
Jan 15 at 19:38
add a comment |
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$begingroup$
Do you know the power rule for integrals?
$endgroup$
– Eric Towers
Jan 15 at 19:26
$begingroup$
@EricTowers yes, and when I did that I got $frac{x^{-n+1}}{-n+1}$ from 1 to 0
$endgroup$
– Addison
Jan 15 at 19:28