Mixing
Studying the convergence of the integral $int_0^pi frac{ln(sin(x))}{x}dx$
$begingroup$
Studying the convergence of the integral $$int_0^pi frac{ln(sin(x))}{x}dx$$
I'm studying for an exam in calculus and I saw this problem and I'm having trouble showing that this integral diverges (That's what says in the solution )
I tried separating the integral like this $$int_0^pi frac{ln(sin(x))}{x}dx=int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx+int_frac{pi}{2}^pifrac{ln(sin(x))}{x}dx$$
Then I substituted $x=pi-t$ for the second integral and I get
$$int_0^pi frac{ln(sin(x))}{x}dx=piint_0^{frac{pi}{2}}frac{ln(sin(x))}{x(pi-x)}dx$$
But I don't know how to procede.
I also tried a different method with the limit test but I don't know with what function to compare. Can someone help me with this problem?
calculus improper-integrals
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
$endgroup$
|
show 2 more comments
$begingroup$
Studying the convergence of the integral $$int_0^pi frac{ln(sin(x))}{x}dx$$
I'm studying for an exam in calculus and I saw this problem and I'm having trouble showing that this integral diverges (That's what says in the solution )
I tried separating the integral like this $$int_0^pi frac{ln(sin(x))}{x}dx=int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx+int_frac{pi}{2}^pifrac{ln(sin(x))}{x}dx$$
Then I substituted $x=pi-t$ for the second integral and I get
$$int_0^pi frac{ln(sin(x))}{x}dx=piint_0^{frac{pi}{2}}frac{ln(sin(x))}{x(pi-x)}dx$$
But I don't know how to procede.
I also tried a different method with the limit test but I don't know with what function to compare. Can someone help me with this problem?
calculus improper-integrals
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
$endgroup$
$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51
|
show 2 more comments
$begingroup$
Studying the convergence of the integral $$int_0^pi frac{ln(sin(x))}{x}dx$$
I'm studying for an exam in calculus and I saw this problem and I'm having trouble showing that this integral diverges (That's what says in the solution )
I tried separating the integral like this $$int_0^pi frac{ln(sin(x))}{x}dx=int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx+int_frac{pi}{2}^pifrac{ln(sin(x))}{x}dx$$
Then I substituted $x=pi-t$ for the second integral and I get
$$int_0^pi frac{ln(sin(x))}{x}dx=piint_0^{frac{pi}{2}}frac{ln(sin(x))}{x(pi-x)}dx$$
But I don't know how to procede.
I also tried a different method with the limit test but I don't know with what function to compare. Can someone help me with this problem?
calculus improper-integrals
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
$endgroup$
Studying the convergence of the integral $$int_0^pi frac{ln(sin(x))}{x}dx$$
I'm studying for an exam in calculus and I saw this problem and I'm having trouble showing that this integral diverges (That's what says in the solution )
I tried separating the integral like this $$int_0^pi frac{ln(sin(x))}{x}dx=int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx+int_frac{pi}{2}^pifrac{ln(sin(x))}{x}dx$$
Then I substituted $x=pi-t$ for the second integral and I get
$$int_0^pi frac{ln(sin(x))}{x}dx=piint_0^{frac{pi}{2}}frac{ln(sin(x))}{x(pi-x)}dx$$
But I don't know how to procede.
I also tried a different method with the limit test but I don't know with what function to compare. Can someone help me with this problem?
calculus improper-integrals
calculus improper-integrals
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
asked Jan 27 at 15:29
J.DaneJ.Dane
375214
375214
$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51
|
show 2 more comments
$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51
$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
$$
doesn't converge.
The integrand is negative and we know $sin x < x$ for $x>0$. Using the fact that $ln x$ is increasing, we have
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
leq int_0^{frac{pi}{2}}frac{ln(x)}{x}dx
= int_{-infty}^{ln pi/2} t , dt.
$$
The last integral obviously diverges to $-infty$.
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
$$
doesn't converge.
The integrand is negative and we know $sin x < x$ for $x>0$. Using the fact that $ln x$ is increasing, we have
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
leq int_0^{frac{pi}{2}}frac{ln(x)}{x}dx
= int_{-infty}^{ln pi/2} t , dt.
$$
The last integral obviously diverges to $-infty$.
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
add a comment |
$begingroup$
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
$$
doesn't converge.
The integrand is negative and we know $sin x < x$ for $x>0$. Using the fact that $ln x$ is increasing, we have
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
leq int_0^{frac{pi}{2}}frac{ln(x)}{x}dx
= int_{-infty}^{ln pi/2} t , dt.
$$
The last integral obviously diverges to $-infty$.
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
add a comment |
$begingroup$
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
$$
doesn't converge.
The integrand is negative and we know $sin x < x$ for $x>0$. Using the fact that $ln x$ is increasing, we have
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
leq int_0^{frac{pi}{2}}frac{ln(x)}{x}dx
= int_{-infty}^{ln pi/2} t , dt.
$$
The last integral obviously diverges to $-infty$.
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
$endgroup$
I assume it's an improper integral (from both sides). So let's split it into two integrals as you did and let's show that the first integral on RHS
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
$$
doesn't converge.
The integrand is negative and we know $sin x < x$ for $x>0$. Using the fact that $ln x$ is increasing, we have
$$
int_0^{frac{pi}{2}}frac{ln(sin(x))}{x}dx
leq int_0^{frac{pi}{2}}frac{ln(x)}{x}dx
= int_{-infty}^{ln pi/2} t , dt.
$$
The last integral obviously diverges to $-infty$.
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
answered Jan 27 at 16:06
Roman HricRoman Hric
32115
32115
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
add a comment |
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
Thank you very much for this simple answer.
$endgroup$
– J.Dane
Jan 27 at 16:12
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
You're welcome. Of course, if it's done in detail, one should write improper integrals as limits of proper ones.
$endgroup$
– Roman Hric
Jan 27 at 16:16
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
$begingroup$
Yeah yeah, of course, I know that
$endgroup$
– J.Dane
Jan 27 at 19:50
add a comment |
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$begingroup$
The function is not bounded in the interval 0 to pi. You cannot talk about the Riemann integral in the first place.
$endgroup$
– John Mitchell
Jan 27 at 15:31
$begingroup$
I think your integral does not converge
$endgroup$
– Dr. Sonnhard Graubner
Jan 27 at 15:36
$begingroup$
I know it doesn't but I don't know how to prove it
$endgroup$
– J.Dane
Jan 27 at 15:37
$begingroup$
@Dr.SonnhardGraubner That's the whole (stated) point of the question: "I'm having trouble showing that this integral diverges"
$endgroup$
– Clement C.
Jan 27 at 15:50
$begingroup$
@J.Dane Are you talking about Lebesgue integral? Also, do you know how to show (say) that $int_0^1 frac{ln x}{x}dx$ diverges?
$endgroup$
– Clement C.
Jan 27 at 15:51