Improper integral convergence $I = int_{0^+}^{1^-}frac{log(x)}{1-x}dx$
$begingroup$
I was solving a few problems regarding convergence and divergence when I ran into this one. I tried searching on the internet but couldn't find an exact match to the problem. The task is to determine whether the following integral diverges or converges.
$$I = int_{0^+}^{1^-}frac{log(x)}{1-x}dx$$
I tried solving it using comparison test but could not find an appropriate function to test it against. No other tests would fit. I also tried integrating the function using by parts but only reached till here (not entirely sure whether it is right or wrong) ($log(x)$ - first part and $(1-x)$ - second part)
$$I=-log(x)cdot log(1-x) + int frac{log(1-x)}{x}dx$$
Putting $y=1-x$ in second part makes it equal to the original integral. So
$$I=-frac12log(x)cdot log(1-x)$$
Here, I face problems putting in the limits.
Please help.
convergence improper-integrals divergent-series
edited Jan 10 at 16:17
Robert Z
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
I was solving a few problems regarding convergence and divergence when I ran into this one. I tried searching on the internet but couldn't find an exact match to the problem. The task is to determine whether the following integral diverges or converges.
$$I = int_{0^+}^{1^-}frac{log(x)}{1-x}dx$$
I tried solving it using comparison test but could not find an appropriate function to test it against. No other tests would fit. I also tried integrating the function using by parts but only reached till here (not entirely sure whether it is right or wrong) ($log(x)$ - first part and $(1-x)$ - second part)
$$I=-log(x)cdot log(1-x) + int frac{log(1-x)}{x}dx$$
Putting $y=1-x$ in second part makes it equal to the original integral. So
$$I=-frac12log(x)cdot log(1-x)$$
Here, I face problems putting in the limits.
Please help.
convergence improper-integrals divergent-series
edited Jan 10 at 16:17
Robert Z
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
$endgroup$
add a comment |
$begingroup$
I was solving a few problems regarding convergence and divergence when I ran into this one. I tried searching on the internet but couldn't find an exact match to the problem. The task is to determine whether the following integral diverges or converges.
$$I = int_{0^+}^{1^-}frac{log(x)}{1-x}dx$$
I tried solving it using comparison test but could not find an appropriate function to test it against. No other tests would fit. I also tried integrating the function using by parts but only reached till here (not entirely sure whether it is right or wrong) ($log(x)$ - first part and $(1-x)$ - second part)
$$I=-log(x)cdot log(1-x) + int frac{log(1-x)}{x}dx$$
Putting $y=1-x$ in second part makes it equal to the original integral. So
$$I=-frac12log(x)cdot log(1-x)$$
Here, I face problems putting in the limits.
Please help.
convergence improper-integrals divergent-series
edited Jan 10 at 16:17
Robert Z
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
$endgroup$
I was solving a few problems regarding convergence and divergence when I ran into this one. I tried searching on the internet but couldn't find an exact match to the problem. The task is to determine whether the following integral diverges or converges.
$$I = int_{0^+}^{1^-}frac{log(x)}{1-x}dx$$
I tried solving it using comparison test but could not find an appropriate function to test it against. No other tests would fit. I also tried integrating the function using by parts but only reached till here (not entirely sure whether it is right or wrong) ($log(x)$ - first part and $(1-x)$ - second part)
$$I=-log(x)cdot log(1-x) + int frac{log(1-x)}{x}dx$$
Putting $y=1-x$ in second part makes it equal to the original integral. So
$$I=-frac12log(x)cdot log(1-x)$$
Here, I face problems putting in the limits.
Please help.
convergence improper-integrals divergent-series
convergence improper-integrals divergent-series
edited Jan 10 at 16:17
Robert Z
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
edited Jan 10 at 16:17
Robert Z
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
edited Jan 10 at 16:17
Robert Z
101k1072145
edited Jan 10 at 16:17
Robert Z
101k1072145
edited Jan 10 at 16:17
Robert Z
101k1072145
101k1072145
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
asked Jan 10 at 16:09
Sauhard SharmaSauhard Sharma
953318
953318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. The function $frac{log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $xto 0^+$ then
$$frac{log(x)}{1-x}sim log(x)$$
Is $log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $xto 1^-$,
$$frac{log(x)}{1-x}=frac{log(1-(1-x))}{1-x}sim frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see
integral representations of $zeta(2)$.
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Hint. The function $frac{log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $xto 0^+$ then
$$frac{log(x)}{1-x}sim log(x)$$
Is $log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $xto 1^-$,
$$frac{log(x)}{1-x}=frac{log(1-(1-x))}{1-x}sim frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see
integral representations of $zeta(2)$.
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
add a comment |
$begingroup$
Hint. The function $frac{log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $xto 0^+$ then
$$frac{log(x)}{1-x}sim log(x)$$
Is $log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $xto 1^-$,
$$frac{log(x)}{1-x}=frac{log(1-(1-x))}{1-x}sim frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see
integral representations of $zeta(2)$.
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
$endgroup$
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
add a comment |
$begingroup$
Hint. The function $frac{log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $xto 0^+$ then
$$frac{log(x)}{1-x}sim log(x)$$
Is $log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $xto 1^-$,
$$frac{log(x)}{1-x}=frac{log(1-(1-x))}{1-x}sim frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see
integral representations of $zeta(2)$.
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
$endgroup$
Hint. The function $frac{log(x)}{1-x}$ is continuous and negative in $(0,1)$ so we can apply the comparison test.
As $xto 0^+$ then
$$frac{log(x)}{1-x}sim log(x)$$
Is $log(x)$ integrable in a right neighbourhood of $0$?
Moreover, as $xto 1^-$,
$$frac{log(x)}{1-x}=frac{log(1-(1-x))}{1-x}sim frac{-(1-x)}{1-x}=-1.$$
What may we conclude?
P.S. The value of such integral is quite famous: see
integral representations of $zeta(2)$.
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
edited Jan 10 at 16:19
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
answered Jan 10 at 16:14
Robert ZRobert Z
101k1072145
101k1072145
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
add a comment |
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
Any further doubt?
$endgroup$
– Robert Z
Jan 10 at 16:56
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
$log(x)$ is integrable in the neighbourhood of $0$. This means that the function is finite at left end. Also, the right end limit exists and is finite. This means that the function is finite everywhere meaning the area of the function is also finite. That implies that the function converges. Right ?
$endgroup$
– Sauhard Sharma
Jan 10 at 17:36
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
No, the function is NOT finite at left end (it goes to $-infty$), but its integral is finite. Hence the integral is convergent.
$endgroup$
– Robert Z
Jan 10 at 17:49
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
$begingroup$
I understand now. Thanks so much !!!
$endgroup$
– Sauhard Sharma
Jan 10 at 18:14
add a comment |
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