What is the value of the integral $int_0^infty 0dx$?
$begingroup$
I know that this question might be very simple, but I want to know: what is the value of following integral?
$$int_0^infty0dx$$
calculus integration improper-integrals
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
$endgroup$
add a comment |
$begingroup$
I know that this question might be very simple, but I want to know: what is the value of following integral?
$$int_0^infty0dx$$
calculus integration improper-integrals
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
$endgroup$
add a comment |
$begingroup$
I know that this question might be very simple, but I want to know: what is the value of following integral?
$$int_0^infty0dx$$
calculus integration improper-integrals
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
$endgroup$
I know that this question might be very simple, but I want to know: what is the value of following integral?
$$int_0^infty0dx$$
calculus integration improper-integrals
calculus integration improper-integrals
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
edited Dec 29 '18 at 20:51
Eevee Trainer
8,32421439
8,32421439
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
asked Dec 29 '18 at 20:38
blindProgrammerblindProgrammer
1287
1287
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The trick with such things is always: look at the definition, and just apply it carefully.
By definition, $int_0^infty f(x)dx$ is just $lim_{rrightarrowinfty} int_0^r f(x)dx$, so we just need to calculate $int_0^r 0dx$ for an arbitrary real $r$.
But $int_0^r 0dx$ is always $0$ (you should already be comfortable with this).
So $int_0^infty f(x)dx=lim_{rrightarrowinfty} int_0^r f(x)dx=lim_{rrightarrowinfty}0=0$.
The reason $int_0^infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $int_0^infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
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add a comment |
$begingroup$
Hint:
$$forall t>0: int_0^t 0 mathrm{d}x=0$$
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
$endgroup$
add a comment |
$begingroup$
A more intuitive angle: recall that
$$int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b to infty,$ and $f(x)=0$. Then we have
$$int_0^infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $xgeq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
$endgroup$
add a comment |
$begingroup$
By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The trick with such things is always: look at the definition, and just apply it carefully.
By definition, $int_0^infty f(x)dx$ is just $lim_{rrightarrowinfty} int_0^r f(x)dx$, so we just need to calculate $int_0^r 0dx$ for an arbitrary real $r$.
But $int_0^r 0dx$ is always $0$ (you should already be comfortable with this).
So $int_0^infty f(x)dx=lim_{rrightarrowinfty} int_0^r f(x)dx=lim_{rrightarrowinfty}0=0$.
The reason $int_0^infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $int_0^infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
$endgroup$
add a comment |
$begingroup$
The trick with such things is always: look at the definition, and just apply it carefully.
By definition, $int_0^infty f(x)dx$ is just $lim_{rrightarrowinfty} int_0^r f(x)dx$, so we just need to calculate $int_0^r 0dx$ for an arbitrary real $r$.
But $int_0^r 0dx$ is always $0$ (you should already be comfortable with this).
So $int_0^infty f(x)dx=lim_{rrightarrowinfty} int_0^r f(x)dx=lim_{rrightarrowinfty}0=0$.
The reason $int_0^infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $int_0^infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
$endgroup$
add a comment |
$begingroup$
The trick with such things is always: look at the definition, and just apply it carefully.
By definition, $int_0^infty f(x)dx$ is just $lim_{rrightarrowinfty} int_0^r f(x)dx$, so we just need to calculate $int_0^r 0dx$ for an arbitrary real $r$.
But $int_0^r 0dx$ is always $0$ (you should already be comfortable with this).
So $int_0^infty f(x)dx=lim_{rrightarrowinfty} int_0^r f(x)dx=lim_{rrightarrowinfty}0=0$.
The reason $int_0^infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $int_0^infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
$endgroup$
The trick with such things is always: look at the definition, and just apply it carefully.
By definition, $int_0^infty f(x)dx$ is just $lim_{rrightarrowinfty} int_0^r f(x)dx$, so we just need to calculate $int_0^r 0dx$ for an arbitrary real $r$.
But $int_0^r 0dx$ is always $0$ (you should already be comfortable with this).
So $int_0^infty f(x)dx=lim_{rrightarrowinfty} int_0^r f(x)dx=lim_{rrightarrowinfty}0=0$.
The reason $int_0^infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $int_0^infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
answered Dec 29 '18 at 20:41
Noah SchweberNoah Schweber
127k10151291
127k10151291
add a comment |
add a comment |
$begingroup$
Hint:
$$forall t>0: int_0^t 0 mathrm{d}x=0$$
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
$endgroup$
add a comment |
$begingroup$
Hint:
$$forall t>0: int_0^t 0 mathrm{d}x=0$$
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
$endgroup$
add a comment |
$begingroup$
Hint:
$$forall t>0: int_0^t 0 mathrm{d}x=0$$
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
$endgroup$
Hint:
$$forall t>0: int_0^t 0 mathrm{d}x=0$$
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
answered Dec 29 '18 at 20:40
BotondBotond
6,44831034
6,44831034
add a comment |
add a comment |
$begingroup$
A more intuitive angle: recall that
$$int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b to infty,$ and $f(x)=0$. Then we have
$$int_0^infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $xgeq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
$endgroup$
add a comment |
$begingroup$
A more intuitive angle: recall that
$$int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b to infty,$ and $f(x)=0$. Then we have
$$int_0^infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $xgeq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
$endgroup$
add a comment |
$begingroup$
A more intuitive angle: recall that
$$int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b to infty,$ and $f(x)=0$. Then we have
$$int_0^infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $xgeq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
$endgroup$
A more intuitive angle: recall that
$$int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b to infty,$ and $f(x)=0$. Then we have
$$int_0^infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $xgeq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
answered Dec 29 '18 at 20:49
Eevee TrainerEevee Trainer
8,32421439
8,32421439
add a comment |
add a comment |
$begingroup$
By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
$endgroup$
add a comment |
$begingroup$
By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
$endgroup$
add a comment |
$begingroup$
By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
$endgroup$
By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
answered Dec 29 '18 at 21:27
Bernard MasséBernard Massé
424145
424145
add a comment |
add a comment |
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