A continuous function $f$ verifies : $int_0^+infty f(x).dx= +infty$. $int_-infty^0 f(x).dx= -infty$.

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0
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In general,for a continuous function $f$ verifies :
$int_0^+infty f(x).dx= +infty$.
$int_-infty^0 f(x).dx= -infty$.
What can we say about $int_-infty^+infty f(x).dx$.
I took the example of $f(x)=2x$ it verifies the hypothesis,
and I've found that :
$int_-infty^+infty 2xdx$ doesn't exit !
I think there are examples where the integral may exist ! (like a finite number, or infinity).




real-analysis integration improper-integrals




share|cite|improve this question

















  • 3




    $int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
    – Botond
    Jul 31 at 22:32











  • Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
    – Bouali Anas
    Jul 31 at 22:34










  • Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
    – Rumpelstiltskin
    Jul 31 at 22:34










  • there's a diffrence between equals $infty$ and doesnt exist !
    – Bouali Anas
    Jul 31 at 22:36










  • If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
    – Arthur
    Jul 31 at 22:44














up vote
0
down vote

favorite












In general,for a continuous function $f$ verifies :
$int_0^+infty f(x).dx= +infty$.
$int_-infty^0 f(x).dx= -infty$.
What can we say about $int_-infty^+infty f(x).dx$.
I took the example of $f(x)=2x$ it verifies the hypothesis,
and I've found that :
$int_-infty^+infty 2xdx$ doesn't exit !
I think there are examples where the integral may exist ! (like a finite number, or infinity).




real-analysis integration improper-integrals




share|cite|improve this question

















  • 3




    $int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
    – Botond
    Jul 31 at 22:32











  • Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
    – Bouali Anas
    Jul 31 at 22:34










  • Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
    – Rumpelstiltskin
    Jul 31 at 22:34










  • there's a diffrence between equals $infty$ and doesnt exist !
    – Bouali Anas
    Jul 31 at 22:36










  • If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
    – Arthur
    Jul 31 at 22:44












up vote
0
down vote

favorite









up vote
0
down vote

favorite











In general,for a continuous function $f$ verifies :
$int_0^+infty f(x).dx= +infty$.
$int_-infty^0 f(x).dx= -infty$.
What can we say about $int_-infty^+infty f(x).dx$.
I took the example of $f(x)=2x$ it verifies the hypothesis,
and I've found that :
$int_-infty^+infty 2xdx$ doesn't exit !
I think there are examples where the integral may exist ! (like a finite number, or infinity).




real-analysis integration improper-integrals




share|cite|improve this question













In general,for a continuous function $f$ verifies :
$int_0^+infty f(x).dx= +infty$.
$int_-infty^0 f(x).dx= -infty$.
What can we say about $int_-infty^+infty f(x).dx$.
I took the example of $f(x)=2x$ it verifies the hypothesis,
and I've found that :
$int_-infty^+infty 2xdx$ doesn't exit !
I think there are examples where the integral may exist ! (like a finite number, or infinity).





real-analysis integration improper-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 22:32









Bernard

110k635102




110k635102









asked Jul 31 at 22:28









Bouali Anas

125




125







  • 3




    $int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
    – Botond
    Jul 31 at 22:32











  • Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
    – Bouali Anas
    Jul 31 at 22:34










  • Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
    – Rumpelstiltskin
    Jul 31 at 22:34










  • there's a diffrence between equals $infty$ and doesnt exist !
    – Bouali Anas
    Jul 31 at 22:36










  • If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
    – Arthur
    Jul 31 at 22:44












  • 3




    $int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
    – Botond
    Jul 31 at 22:32











  • Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
    – Bouali Anas
    Jul 31 at 22:34










  • Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
    – Rumpelstiltskin
    Jul 31 at 22:34










  • there's a diffrence between equals $infty$ and doesnt exist !
    – Bouali Anas
    Jul 31 at 22:36










  • If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
    – Arthur
    Jul 31 at 22:44







3




3




$int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
– Botond
Jul 31 at 22:32





$int_-infty^inftyf$ exists if both $int_-infty^0f$ and $int_0^inftyf$ exists (and finite). But they are not finite in your case.
– Botond
Jul 31 at 22:32













Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
– Bouali Anas
Jul 31 at 22:34




Yeah that's, the problems, there is some example where the integral in $mathbbR$ equals $infty$ ?
– Bouali Anas
Jul 31 at 22:34












Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
– Rumpelstiltskin
Jul 31 at 22:34




Integral itself doesn't exists, but see this wikipedia link, that makes sense of those kind of integrals.
– Rumpelstiltskin
Jul 31 at 22:34












there's a diffrence between equals $infty$ and doesnt exist !
– Bouali Anas
Jul 31 at 22:36




there's a diffrence between equals $infty$ and doesnt exist !
– Bouali Anas
Jul 31 at 22:36












If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
– Arthur
Jul 31 at 22:44




If the $int_-infty^0$ is $-infty$, then $int_-infty^infty$ cannot be $infty$. It can be undefined, or $-infty$, but not $infty$.
– Arthur
Jul 31 at 22:44










1 Answer
1






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If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $mathbbR$ is 0. So, $$lim_arightarrow infty int_-a^a f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.



To get the PV to equal some $bin mathbbR$ you can take an odd function $f$ and manipulate it to a function $$
g(x)=
begincases
h(x) quadtextforquad xin [x_0,x_1],\
f(x) quad textelse,
endcases
$$
where $$int_x_0^x_1 h(x) dx=int_x_0^x_1 f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$






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  • Thank you, that really answer my question !
    – Bouali Anas
    Aug 1 at 10:18










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $mathbbR$ is 0. So, $$lim_arightarrow infty int_-a^a f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.



To get the PV to equal some $bin mathbbR$ you can take an odd function $f$ and manipulate it to a function $$
g(x)=
begincases
h(x) quadtextforquad xin [x_0,x_1],\
f(x) quad textelse,
endcases
$$
where $$int_x_0^x_1 h(x) dx=int_x_0^x_1 f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$






share|cite|improve this answer





















  • Thank you, that really answer my question !
    – Bouali Anas
    Aug 1 at 10:18














up vote
2
down vote



accepted










If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $mathbbR$ is 0. So, $$lim_arightarrow infty int_-a^a f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.



To get the PV to equal some $bin mathbbR$ you can take an odd function $f$ and manipulate it to a function $$
g(x)=
begincases
h(x) quadtextforquad xin [x_0,x_1],\
f(x) quad textelse,
endcases
$$
where $$int_x_0^x_1 h(x) dx=int_x_0^x_1 f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$






share|cite|improve this answer





















  • Thank you, that really answer my question !
    – Bouali Anas
    Aug 1 at 10:18












up vote
2
down vote



accepted







up vote
2
down vote



accepted






If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $mathbbR$ is 0. So, $$lim_arightarrow infty int_-a^a f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.



To get the PV to equal some $bin mathbbR$ you can take an odd function $f$ and manipulate it to a function $$
g(x)=
begincases
h(x) quadtextforquad xin [x_0,x_1],\
f(x) quad textelse,
endcases
$$
where $$int_x_0^x_1 h(x) dx=int_x_0^x_1 f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$






share|cite|improve this answer













If principal values are good enough for you, you can say that for any integrable odd function $f$ the integral over $mathbbR$ is 0. So, $$lim_arightarrow infty int_-a^a f(x) dx=0.$$ Then of course you can generate a simple example where the PV is any number.



To get the PV to equal some $bin mathbbR$ you can take an odd function $f$ and manipulate it to a function $$
g(x)=
begincases
h(x) quadtextforquad xin [x_0,x_1],\
f(x) quad textelse,
endcases
$$
where $$int_x_0^x_1 h(x) dx=int_x_0^x_1 f(x) dx+b.$$ And if you want $g$ to be continious you must have $h(x_i)=f(x_i),i=0,1.$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 31 at 23:08









ty.

38519




38519











  • Thank you, that really answer my question !
    – Bouali Anas
    Aug 1 at 10:18
















  • Thank you, that really answer my question !
    – Bouali Anas
    Aug 1 at 10:18















Thank you, that really answer my question !
– Bouali Anas
Aug 1 at 10:18




Thank you, that really answer my question !
– Bouali Anas
Aug 1 at 10:18












 

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