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Proving that an improper integral converges
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Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Prove that the improper integral $int_0^1g(x)dx$ converges, that the series $sum_k=1^inftyfrac1k^2=1+frac14+frac19+cdots$ converges, and that the two limits are equal.
My attempt of a proof so far is the following:
First I defined $f(x)=int_0^xfrac1ulnfrac11-udu$ and performed a Taylor expansion at 0 to get $f(x)=int_0^xfrac1ulnfrac11-udu=int_0^1sum_k=1^inftyfracu^k-1k=sum_k=1^inftyfracx^kk^2$ for $|x|<1$. This part of the proof I had trouble with, but eventually worked through it with some help. Now I need some more help.
The improper integral $int_0^1g(x)dx$ can be expressed as $$lim_xto1^-f(x)=lim_tto1^-int_0^tfrac1xlnfrac11-xdx=lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2.$$
It's now shown that the improper integral equals the series, so now all I need to do is show that the series converges and the proof will be satisfied. I will use the integral test for $sum_k=1^inftyfrac1k^2$. Let $f(x)=frac1x^2$. $f(x)$ is continuous and positive for all $xgeq1$ and $f'(x)=-frac2x^3$ so $f(x)$ is decreasing for all $xgeq1$. The conditions for the integral test are satisfied. The integral test for convergence is $$int_1^inftyfrac1x^2dx=lim_btoinftyint_1^bfrac1x^2dx=lim_btoinftybigg[-frac1xbigg]_1^b=[0-(-1)]=1.$$
The integral converges so, therefore, the series must converge as well via the integral test. Thus, $int_0^1g(x)dx$ and the series $sum_k=1^inftyfrac1k^2$ both converge and are equal.
Are there any flaws in my proof and where can I improve?
real-analysis proof-verification
asked Jul 26 at 19:38
Peetrius
358111
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up vote
2
down vote
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Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Prove that the improper integral $int_0^1g(x)dx$ converges, that the series $sum_k=1^inftyfrac1k^2=1+frac14+frac19+cdots$ converges, and that the two limits are equal.
My attempt of a proof so far is the following:
First I defined $f(x)=int_0^xfrac1ulnfrac11-udu$ and performed a Taylor expansion at 0 to get $f(x)=int_0^xfrac1ulnfrac11-udu=int_0^1sum_k=1^inftyfracu^k-1k=sum_k=1^inftyfracx^kk^2$ for $|x|<1$. This part of the proof I had trouble with, but eventually worked through it with some help. Now I need some more help.
The improper integral $int_0^1g(x)dx$ can be expressed as $$lim_xto1^-f(x)=lim_tto1^-int_0^tfrac1xlnfrac11-xdx=lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2.$$
It's now shown that the improper integral equals the series, so now all I need to do is show that the series converges and the proof will be satisfied. I will use the integral test for $sum_k=1^inftyfrac1k^2$. Let $f(x)=frac1x^2$. $f(x)$ is continuous and positive for all $xgeq1$ and $f'(x)=-frac2x^3$ so $f(x)$ is decreasing for all $xgeq1$. The conditions for the integral test are satisfied. The integral test for convergence is $$int_1^inftyfrac1x^2dx=lim_btoinftyint_1^bfrac1x^2dx=lim_btoinftybigg[-frac1xbigg]_1^b=[0-(-1)]=1.$$
The integral converges so, therefore, the series must converge as well via the integral test. Thus, $int_0^1g(x)dx$ and the series $sum_k=1^inftyfrac1k^2$ both converge and are equal.
Are there any flaws in my proof and where can I improve?
real-analysis proof-verification
asked Jul 26 at 19:38
Peetrius
358111
1
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
2
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
2
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
1
@MarkViola: Good point.
– RRL
Jul 26 at 19:58
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Prove that the improper integral $int_0^1g(x)dx$ converges, that the series $sum_k=1^inftyfrac1k^2=1+frac14+frac19+cdots$ converges, and that the two limits are equal.
My attempt of a proof so far is the following:
First I defined $f(x)=int_0^xfrac1ulnfrac11-udu$ and performed a Taylor expansion at 0 to get $f(x)=int_0^xfrac1ulnfrac11-udu=int_0^1sum_k=1^inftyfracu^k-1k=sum_k=1^inftyfracx^kk^2$ for $|x|<1$. This part of the proof I had trouble with, but eventually worked through it with some help. Now I need some more help.
The improper integral $int_0^1g(x)dx$ can be expressed as $$lim_xto1^-f(x)=lim_tto1^-int_0^tfrac1xlnfrac11-xdx=lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2.$$
It's now shown that the improper integral equals the series, so now all I need to do is show that the series converges and the proof will be satisfied. I will use the integral test for $sum_k=1^inftyfrac1k^2$. Let $f(x)=frac1x^2$. $f(x)$ is continuous and positive for all $xgeq1$ and $f'(x)=-frac2x^3$ so $f(x)$ is decreasing for all $xgeq1$. The conditions for the integral test are satisfied. The integral test for convergence is $$int_1^inftyfrac1x^2dx=lim_btoinftyint_1^bfrac1x^2dx=lim_btoinftybigg[-frac1xbigg]_1^b=[0-(-1)]=1.$$
The integral converges so, therefore, the series must converge as well via the integral test. Thus, $int_0^1g(x)dx$ and the series $sum_k=1^inftyfrac1k^2$ both converge and are equal.
Are there any flaws in my proof and where can I improve?
real-analysis proof-verification
asked Jul 26 at 19:38
Peetrius
358111
Consider the function $$g(x)=begincases frac1xlnfrac11-xhspace.5cm xneq0,\ 1 hspace1.8cm x=0.endcases$$
Prove that the improper integral $int_0^1g(x)dx$ converges, that the series $sum_k=1^inftyfrac1k^2=1+frac14+frac19+cdots$ converges, and that the two limits are equal.
My attempt of a proof so far is the following:
First I defined $f(x)=int_0^xfrac1ulnfrac11-udu$ and performed a Taylor expansion at 0 to get $f(x)=int_0^xfrac1ulnfrac11-udu=int_0^1sum_k=1^inftyfracu^k-1k=sum_k=1^inftyfracx^kk^2$ for $|x|<1$. This part of the proof I had trouble with, but eventually worked through it with some help. Now I need some more help.
The improper integral $int_0^1g(x)dx$ can be expressed as $$lim_xto1^-f(x)=lim_tto1^-int_0^tfrac1xlnfrac11-xdx=lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2.$$
It's now shown that the improper integral equals the series, so now all I need to do is show that the series converges and the proof will be satisfied. I will use the integral test for $sum_k=1^inftyfrac1k^2$. Let $f(x)=frac1x^2$. $f(x)$ is continuous and positive for all $xgeq1$ and $f'(x)=-frac2x^3$ so $f(x)$ is decreasing for all $xgeq1$. The conditions for the integral test are satisfied. The integral test for convergence is $$int_1^inftyfrac1x^2dx=lim_btoinftyint_1^bfrac1x^2dx=lim_btoinftybigg[-frac1xbigg]_1^b=[0-(-1)]=1.$$
The integral converges so, therefore, the series must converge as well via the integral test. Thus, $int_0^1g(x)dx$ and the series $sum_k=1^inftyfrac1k^2$ both converge and are equal.
Are there any flaws in my proof and where can I improve?
real-analysis proof-verification
asked Jul 26 at 19:38
Peetrius
358111
asked Jul 26 at 19:38
Peetrius
358111
asked Jul 26 at 19:38
Peetrius
358111
asked Jul 26 at 19:38
Peetrius
358111
358111
1
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
2
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
2
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
1
@MarkViola: Good point.
– RRL
Jul 26 at 19:58
add a comment |Â
1
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
2
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
2
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
1
@MarkViola: Good point.
– RRL
Jul 26 at 19:58
1
1
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
2
2
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
2
2
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
1
1
@MarkViola: Good point.
– RRL
Jul 26 at 19:58
@MarkViola: Good point.
– RRL
Jul 26 at 19:58
add a comment |Â
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1
It should be enough to show that $int_0^1 log(1-x),dx$ exists as an improper Riemann integral.
– Mark Viola
Jul 26 at 19:43
2
This works. You do need to justify $lim_xto1^-sum_k=1^inftyfracx^kk^2=sum_k=1^inftyfrac1k^2$ which you can with Abel's theorem
– RRL
Jul 26 at 19:50
2
@RRL Another way is to recognize that $sum_k=1^infty fracx^kk^2$ converges uniformly on $[-1,1]$ and hence is continuous there.
– Mark Viola
Jul 26 at 19:56
1
@MarkViola: Good point.
– RRL
Jul 26 at 19:58