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Can we use symmetry rules in improper integrals?
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I wish to evaluate the integral $$I=int^infty_-inftyxe^-x^2dx$$
Can I simply note that that $f(x)=xe^-x^2$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $mathbbR$," it tempts me to accept the symmetry argument.
The actual answer via limits is $0$.
improper-integrals
asked Jul 16 at 8:58
coreyman317
687218
add a comment |Â
up vote
28
down vote
favorite
I wish to evaluate the integral $$I=int^infty_-inftyxe^-x^2dx$$
Can I simply note that that $f(x)=xe^-x^2$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $mathbbR$," it tempts me to accept the symmetry argument.
The actual answer via limits is $0$.
improper-integrals
asked Jul 16 at 8:58
coreyman317
687218
3
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
1
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23
add a comment |Â
up vote
28
down vote
favorite
up vote
28
down vote
favorite
I wish to evaluate the integral $$I=int^infty_-inftyxe^-x^2dx$$
Can I simply note that that $f(x)=xe^-x^2$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $mathbbR$," it tempts me to accept the symmetry argument.
The actual answer via limits is $0$.
improper-integrals
asked Jul 16 at 8:58
coreyman317
687218
I wish to evaluate the integral $$I=int^infty_-inftyxe^-x^2dx$$
Can I simply note that that $f(x)=xe^-x^2$ is an odd function and say $I=0$? The only reason I have doubts is because of assuming the two infinities have the same length. However, when I hear people say, "...the integral of an odd function vanishes on $mathbbR$," it tempts me to accept the symmetry argument.
The actual answer via limits is $0$.
improper-integrals
asked Jul 16 at 8:58
coreyman317
687218
asked Jul 16 at 8:58
coreyman317
687218
asked Jul 16 at 8:58
coreyman317
687218
asked Jul 16 at 8:58
coreyman317
687218
687218
3
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
1
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23
add a comment |Â
3
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
1
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23
3
3
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
1
1
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
42
down vote
accepted
Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce†that, say, $int_-infty^+inftyx,mathrm dx=0$.
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
 |Â
show 2 more comments
up vote
27
down vote
One needs to be careful what $$int_-infty^infty f(x),mathrmdx tag1$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)colonmathbb R^2tomathbb R$ be defined as $$F(a,b) = int_a^bf(x),mathrm d x$$ and then define $(1)$ to be $$lim_(a,b)to(-infty,infty)F(a,b). tag2$$
Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $mathbb R^2$ such that $a(t)to -infty$ and $b(t)toinfty$ when $ttoinfty$ and you will have $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_a(t)^b(t) f(x),mathrmdx.$$
In particular, take $a(t) = t$ and $b(t) = -t$ to get $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_-t^t f(x),mathrmdx. tag3$$
You can use symmetry argument on $(3)$ to get what you want.
The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $int_-infty^infty x,mathrm d x$ being the main such counterexample.
answered Jul 16 at 9:45
Ennar
13k32343
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
add a comment |Â
up vote
2
down vote
Well, the primordial counterexample is $int_-infty^infty frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".
Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.
add a comment |Â
up vote
1
down vote
Since
$$
int_-infty^inftyleft|,xe^-x^2right|,mathrmdx=1
$$
we have that $xe^-x^2in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $int_-infty^inftyfracx,mathrmdxx^2+1$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.
answered Jul 19 at 23:32
robjohn♦
258k26297612
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
42
down vote
accepted
Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce†that, say, $int_-infty^+inftyx,mathrm dx=0$.
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
 |Â
show 2 more comments
up vote
42
down vote
accepted
Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce†that, say, $int_-infty^+inftyx,mathrm dx=0$.
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
 |Â
show 2 more comments
up vote
42
down vote
accepted
up vote
42
down vote
accepted
Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce†that, say, $int_-infty^+inftyx,mathrm dx=0$.
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
Yes, you can use that symmetry argument for improper integrals, but only after you proved that the integral exists (which is the case in your example). Otherwise, you might “deduce†that, say, $int_-infty^+inftyx,mathrm dx=0$.
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
edited Jul 19 at 23:06
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
answered Jul 16 at 9:01
José Carlos Santos
114k1698177
114k1698177
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
 |Â
show 2 more comments
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
If a function is odd and vanishes in the limits, is there any case where you can not instantly jump to saying its -inf to inf integral is 0?
– iammax
Jul 16 at 12:00
7
7
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
@iammax Sure, take any non-integrable function and mirror it. For example, $fracsgn(x)$
– Serge Seredenko
Jul 16 at 12:14
1
1
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
Hmm are we using Lebesgue or Riemann integration? Because with Riemann don't we have $int_-infty^+inftyx mathrm dx = lim_b to infty int_-b^b x mathrm dx = 0$?
– Ovi
Jul 16 at 19:21
6
6
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
@Ovi No, we don't.
– José Carlos Santos
Jul 16 at 19:21
5
5
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
@iammax Another example is $frac xx^2+1$. It's not enough to vanish; it has to vanish more quickly than $x^-1$; the integral of $x^-1$ is $ln(x)$, which diverges.
– Acccumulation
Jul 16 at 22:16
 |Â
show 2 more comments
up vote
27
down vote
One needs to be careful what $$int_-infty^infty f(x),mathrmdx tag1$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)colonmathbb R^2tomathbb R$ be defined as $$F(a,b) = int_a^bf(x),mathrm d x$$ and then define $(1)$ to be $$lim_(a,b)to(-infty,infty)F(a,b). tag2$$
Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $mathbb R^2$ such that $a(t)to -infty$ and $b(t)toinfty$ when $ttoinfty$ and you will have $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_a(t)^b(t) f(x),mathrmdx.$$
In particular, take $a(t) = t$ and $b(t) = -t$ to get $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_-t^t f(x),mathrmdx. tag3$$
You can use symmetry argument on $(3)$ to get what you want.
The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $int_-infty^infty x,mathrm d x$ being the main such counterexample.
answered Jul 16 at 9:45
Ennar
13k32343
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
add a comment |Â
up vote
27
down vote
One needs to be careful what $$int_-infty^infty f(x),mathrmdx tag1$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)colonmathbb R^2tomathbb R$ be defined as $$F(a,b) = int_a^bf(x),mathrm d x$$ and then define $(1)$ to be $$lim_(a,b)to(-infty,infty)F(a,b). tag2$$
Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $mathbb R^2$ such that $a(t)to -infty$ and $b(t)toinfty$ when $ttoinfty$ and you will have $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_a(t)^b(t) f(x),mathrmdx.$$
In particular, take $a(t) = t$ and $b(t) = -t$ to get $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_-t^t f(x),mathrmdx. tag3$$
You can use symmetry argument on $(3)$ to get what you want.
The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $int_-infty^infty x,mathrm d x$ being the main such counterexample.
answered Jul 16 at 9:45
Ennar
13k32343
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
add a comment |Â
up vote
27
down vote
up vote
27
down vote
One needs to be careful what $$int_-infty^infty f(x),mathrmdx tag1$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)colonmathbb R^2tomathbb R$ be defined as $$F(a,b) = int_a^bf(x),mathrm d x$$ and then define $(1)$ to be $$lim_(a,b)to(-infty,infty)F(a,b). tag2$$
Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $mathbb R^2$ such that $a(t)to -infty$ and $b(t)toinfty$ when $ttoinfty$ and you will have $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_a(t)^b(t) f(x),mathrmdx.$$
In particular, take $a(t) = t$ and $b(t) = -t$ to get $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_-t^t f(x),mathrmdx. tag3$$
You can use symmetry argument on $(3)$ to get what you want.
The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $int_-infty^infty x,mathrm d x$ being the main such counterexample.
answered Jul 16 at 9:45
Ennar
13k32343
One needs to be careful what $$int_-infty^infty f(x),mathrmdx tag1$$ stands for. Most natural way to define it is as a double limit; let $F(a,b)colonmathbb R^2tomathbb R$ be defined as $$F(a,b) = int_a^bf(x),mathrm d x$$ and then define $(1)$ to be $$lim_(a,b)to(-infty,infty)F(a,b). tag2$$
Existence of the double limit is pretty strong, it means that you can take any parametrized path $(a(t),b(t))$ in $mathbb R^2$ such that $a(t)to -infty$ and $b(t)toinfty$ when $ttoinfty$ and you will have $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_a(t)^b(t) f(x),mathrmdx.$$
In particular, take $a(t) = t$ and $b(t) = -t$ to get $$int_-infty^infty f(x),mathrmdx = lim_ttoinftyint_-t^t f(x),mathrmdx. tag3$$
You can use symmetry argument on $(3)$ to get what you want.
The RHS of $(3)$ is what is called the Cauchy principal value. However, existence of Cauchy principal value does not imply the existence of the limit in $(2)$, $int_-infty^infty x,mathrm d x$ being the main such counterexample.
answered Jul 16 at 9:45
Ennar
13k32343
answered Jul 16 at 9:45
Ennar
13k32343
answered Jul 16 at 9:45
Ennar
13k32343
answered Jul 16 at 9:45
Ennar
13k32343
13k32343
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
add a comment |Â
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
Will it be correct to state $a(t) = -t$ and $b(t) = t$ and let $t rightarrow infty$ or $a(t) = t$ and $b(t) = -t$ and let $t rightarrow -infty$ ?
– BAYMAX
Jul 17 at 7:28
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
@BAYMAX, well, $xmapsto -x$ is a homeomorphism so it doesn't really matter, it's a very common substitution when dealing with limits. But as I've said, existence of double limit implies that limit is path independent, you might use $(a(t),b(t)) = (-t,t^2)$ or even $(-1/t,1/t)$ and let $tto 0^+$.
– Ennar
Jul 17 at 14:27
add a comment |Â
up vote
2
down vote
Well, the primordial counterexample is $int_-infty^infty frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".
Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.
add a comment |Â
up vote
2
down vote
Well, the primordial counterexample is $int_-infty^infty frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".
Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Well, the primordial counterexample is $int_-infty^infty frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".
Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.
Well, the primordial counterexample is $int_-infty^infty frac1x$ where the value of the integral depends on the path taken at 0. Crossing infinitesimally "below" the pole yields a different sign from crossing infinitesimally "above".
Basically, analytic functions allow you to take any path through the complex plane with equal results as long as the different paths do not pass simple poles on different sides: any circular path integral for analytic functions is zero unless it encloses such points.
answered Jul 17 at 8:07
user577667
add a comment |Â
add a comment |Â
up vote
1
down vote
Since
$$
int_-infty^inftyleft|,xe^-x^2right|,mathrmdx=1
$$
we have that $xe^-x^2in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $int_-infty^inftyfracx,mathrmdxx^2+1$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.
answered Jul 19 at 23:32
robjohn♦
258k26297612
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
add a comment |Â
up vote
1
down vote
Since
$$
int_-infty^inftyleft|,xe^-x^2right|,mathrmdx=1
$$
we have that $xe^-x^2in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $int_-infty^inftyfracx,mathrmdxx^2+1$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.
answered Jul 19 at 23:32
robjohn♦
258k26297612
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since
$$
int_-infty^inftyleft|,xe^-x^2right|,mathrmdx=1
$$
we have that $xe^-x^2in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $int_-infty^inftyfracx,mathrmdxx^2+1$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.
answered Jul 19 at 23:32
robjohn♦
258k26297612
Since
$$
int_-infty^inftyleft|,xe^-x^2right|,mathrmdx=1
$$
we have that $xe^-x^2in L^1$. Thus, it is valid to use symmetry.
If the function is not in $L^1$, such as $int_-infty^inftyfracx,mathrmdxx^2+1$, one usually says that the Cauchy Principal Value of the integral is $0$ by symmetry.
answered Jul 19 at 23:32
robjohn♦
258k26297612
answered Jul 19 at 23:32
robjohn♦
258k26297612
answered Jul 19 at 23:32
robjohn♦
258k26297612
answered Jul 19 at 23:32
robjohn♦
258k26297612
258k26297612
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
add a comment |Â
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
What does it mean to say $|xe^-x^2| in L^1$? I know $L^p$ spaces are Banach spaces, but I don’t know what it means to say a function is an element of an $L^p$ space.
– coreyman317
Jul 20 at 1:52
1
1
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
A function $f$ is in an $L^p$ space on $E$ when $$int_E|f(x)|^p,mathrmdxltinfty$$
– robjohn♦
Jul 20 at 3:08
add a comment |Â
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3
I think you can, if $exists J=int_0^infty x e^-x^2$ and $J < infty$.
– Botond
Jul 16 at 9:00
1
You should first separate the integral to 2 integrals with a single singularity/problematic point in each one (for example, separate to $[-infty,0]$ and $[0,infty]$). If each one converges, than your claim that the integrand is odd is enough to say that the answer is 0
– GuySa
Jul 16 at 9:23