Improper integral of a function
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-2
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I want to study the nature of an improper integral. I separate the integral from $0$ to $1$ and from $1$ to $infty$.
The first integral is divergent (I proved that it is equivalent to Riemann integral, which is divergent in the neighborhood of $0$). But I didn't find how I can prove the convergence or divergence of the second one.
The integral can be found below:
$$int_0^{infty}frac{ln(1+x)^2}{sin^2x}e^{-x} text d x$$
integration improper-integrals
edited Nov 21 at 16:47
Aaron Stevens
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
add a comment |
up vote
-2
down vote
favorite
I want to study the nature of an improper integral. I separate the integral from $0$ to $1$ and from $1$ to $infty$.
The first integral is divergent (I proved that it is equivalent to Riemann integral, which is divergent in the neighborhood of $0$). But I didn't find how I can prove the convergence or divergence of the second one.
The integral can be found below:
$$int_0^{infty}frac{ln(1+x)^2}{sin^2x}e^{-x} text d x$$
integration improper-integrals
edited Nov 21 at 16:47
Aaron Stevens
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
2
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
3
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
2
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I want to study the nature of an improper integral. I separate the integral from $0$ to $1$ and from $1$ to $infty$.
The first integral is divergent (I proved that it is equivalent to Riemann integral, which is divergent in the neighborhood of $0$). But I didn't find how I can prove the convergence or divergence of the second one.
The integral can be found below:
$$int_0^{infty}frac{ln(1+x)^2}{sin^2x}e^{-x} text d x$$
integration improper-integrals
edited Nov 21 at 16:47
Aaron Stevens
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
I want to study the nature of an improper integral. I separate the integral from $0$ to $1$ and from $1$ to $infty$.
The first integral is divergent (I proved that it is equivalent to Riemann integral, which is divergent in the neighborhood of $0$). But I didn't find how I can prove the convergence or divergence of the second one.
The integral can be found below:
$$int_0^{infty}frac{ln(1+x)^2}{sin^2x}e^{-x} text d x$$
integration improper-integrals
integration improper-integrals
edited Nov 21 at 16:47
Aaron Stevens
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
edited Nov 21 at 16:47
Aaron Stevens
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
edited Nov 21 at 16:47
Aaron Stevens
219110
edited Nov 21 at 16:47
Aaron Stevens
219110
edited Nov 21 at 16:47
Aaron Stevens
219110
219110
asked Nov 21 at 16:29
Mohammed Mohammed
122
asked Nov 21 at 16:29
Mohammed Mohammed
122
asked Nov 21 at 16:29
Mohammed Mohammed
122
122
2
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
3
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
2
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40
add a comment |
2
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
3
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
2
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40
2
2
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
3
3
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
2
2
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
The integrand has a singularity at every value $x=kpi.$ And it will diverge at each of those values.
answered Nov 21 at 16:34
B. Goddard
18.2k21340
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The integrand has a singularity at every value $x=kpi.$ And it will diverge at each of those values.
answered Nov 21 at 16:34
B. Goddard
18.2k21340
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
add a comment |
up vote
0
down vote
The integrand has a singularity at every value $x=kpi.$ And it will diverge at each of those values.
answered Nov 21 at 16:34
B. Goddard
18.2k21340
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
add a comment |
up vote
0
down vote
up vote
0
down vote
The integrand has a singularity at every value $x=kpi.$ And it will diverge at each of those values.
answered Nov 21 at 16:34
B. Goddard
18.2k21340
The integrand has a singularity at every value $x=kpi.$ And it will diverge at each of those values.
answered Nov 21 at 16:34
B. Goddard
18.2k21340
edited Nov 21 at 21:46
answered Nov 21 at 16:34
B. Goddard
18.2k21340
answered Nov 21 at 16:34
B. Goddard
18.2k21340
answered Nov 21 at 16:34
B. Goddard
18.2k21340
18.2k21340
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
add a comment |
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
2
2
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
No, at $0$ it doesn't diverge. The integrand tends to $1$. The rest is correct though
– Federico
Nov 21 at 16:36
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
at 0 we use the equivalent we have that ln(1+x)² is equivalent to 2x and sin²x is equivanlent to x² and exp(x) is equivalent to 1 so the above integral is equivalent to 2x/x²=2/x and the integral of 2/x from 0 to a real number diverge ( This is when we separate the integral into integrals the first one is from 0 to 1 and the second from 1 to infini
– Mohammed Mohammed
Nov 21 at 16:52
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
@MohammedMohammed Just plot your integrand. You will see that it does not diverge at $0$. Even if it did, what is the point of splitting the integral at $x=1$?
– Aaron Stevens
Nov 21 at 17:38
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
Near $0$, $ln^2(1+x)approx x^2$, so the integral actually has an integrable singularity.
– Zachary
Nov 21 at 21:30
add a comment |
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2
Wow, that's a pretty big picture for such a small formula...
– Federico
Nov 21 at 16:32
3
You are wrong about the divergence near $0$. The integrand is continuous near $0$
– Federico
Nov 21 at 16:35
2
Also, if you claim (wrongly) that it diverges near $0$, what's the point of studying the behaviour elsewhere? You already know that it diverges
– Federico
Nov 21 at 16:37
To add to @Fredrico, what is the point of splitting the integral at $x=1$ if you believe it to diverge at $0$?
– Aaron Stevens
Nov 21 at 17:40