Can the lower limit of $frac{d}{dx} int^x_a f(t)dt = f(x)$ be $-infty$?












1












$begingroup$


I'm self studying math, based on the fundamental theorem of Calculus, $$frac{d}{dx} int^x_a f(t)dt = f(x)$$ can the lower limit be $-infty$?






calculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:03










  • $begingroup$
    @SangchulLee what is the constant equal to?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:18










  • $begingroup$
    Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:21










  • $begingroup$
    if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:59












  • $begingroup$
    Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 2:05


















1












$begingroup$


I'm self studying math, based on the fundamental theorem of Calculus, $$frac{d}{dx} int^x_a f(t)dt = f(x)$$ can the lower limit be $-infty$?






calculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:03










  • $begingroup$
    @SangchulLee what is the constant equal to?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:18










  • $begingroup$
    Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:21










  • $begingroup$
    if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:59












  • $begingroup$
    Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 2:05
















1












1








1


2



$begingroup$


I'm self studying math, based on the fundamental theorem of Calculus, $$frac{d}{dx} int^x_a f(t)dt = f(x)$$ can the lower limit be $-infty$?






calculus







share|cite|improve this question











$endgroup$




I'm self studying math, based on the fundamental theorem of Calculus, $$frac{d}{dx} int^x_a f(t)dt = f(x)$$ can the lower limit be $-infty$?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 4:40









Andrews

3901317




3901317










asked Dec 8 '18 at 0:49









drerDdrerD

1559




1559








  • 3




    $begingroup$
    If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:03










  • $begingroup$
    @SangchulLee what is the constant equal to?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:18










  • $begingroup$
    Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:21










  • $begingroup$
    if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:59












  • $begingroup$
    Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 2:05
















  • 3




    $begingroup$
    If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:03










  • $begingroup$
    @SangchulLee what is the constant equal to?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:18










  • $begingroup$
    Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 1:21










  • $begingroup$
    if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:59












  • $begingroup$
    Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
    $endgroup$
    – Sangchul Lee
    Dec 8 '18 at 2:05










3




3




$begingroup$
If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
$endgroup$
– Sangchul Lee
Dec 8 '18 at 1:03




$begingroup$
If $int_{-infty}^{x_0} f(t) , dt$ is defined for some $x_0$, then write $$ int_{-infty}^{x} f(t) , dt = underbrace{int_{-infty}^{x_0} f(t) , dt}_{text{constant}} + int_{x_0}^{x} f(t) , dt$$ and apply FToC to the second term!
$endgroup$
– Sangchul Lee
Dec 8 '18 at 1:03












$begingroup$
@SangchulLee what is the constant equal to?
$endgroup$
– drerD
Dec 8 '18 at 1:18




$begingroup$
@SangchulLee what is the constant equal to?
$endgroup$
– drerD
Dec 8 '18 at 1:18












$begingroup$
Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
$endgroup$
– Sangchul Lee
Dec 8 '18 at 1:21




$begingroup$
Nothing can be said unless the function $f$ is specified. But as soon as differentiation is concerned, its value is irrelevant for obtaining FToC-type result for $a=-infty$.
$endgroup$
– Sangchul Lee
Dec 8 '18 at 1:21












$begingroup$
if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
$endgroup$
– drerD
Dec 8 '18 at 1:59






$begingroup$
if you can break down the integral into infinite smaller pieces, $sum^{x}_{k=-infty} f(k)$ then wouldn't it be infinity?
$endgroup$
– drerD
Dec 8 '18 at 1:59














$begingroup$
Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
$endgroup$
– Sangchul Lee
Dec 8 '18 at 2:05






$begingroup$
Putting aside the issue that $sum_{k=0}^{infty} f(k)$ is not necessarily related to the improper integral $int_{0}^{infty} f(t) , dt$, are you asking whether $int_{0}^{infty} f(t) , dt$ always diverges? If so, then the answer is definitely no, much like an infinite sum need not necessarily diverge. For instance, $$ int_{0}^{infty} e^{-x} , dx = 1, qquad int_{0}^{infty} e^{-x^2} , dx = frac{sqrt{pi}}{2}, qquad int_{0}^{infty} frac{dx}{1+x^2} = frac{pi}{2}, quad cdots. $$
$endgroup$
– Sangchul Lee
Dec 8 '18 at 2:05












2 Answers
2






active

oldest

votes


















3












$begingroup$

Yes, lower limit can be $-infty$ but only provided the resulting improper integral converges. This only means if you first integrate from $a$ to $x,$ get that answer, and then let $a to -infty,$ that limit exists.



I'm assuming $f(t)$ is defined and continuous (or sufficiently nice) on $(-infty,x).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:20












  • $begingroup$
    @user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
    $endgroup$
    – coffeemath
    Dec 8 '18 at 1:26










  • $begingroup$
    @user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
    $endgroup$
    – eyeballfrog
    Dec 8 '18 at 5:23



















0












$begingroup$

f(x) depeds only on x not on a.



If f(t)=-t then $frac{d}{dx} intlimits^x_a (-t)dt = -frac{x^2}{2}$ and the lower boun of this function is $-infty$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what's lim sup?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:17











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Yes, lower limit can be $-infty$ but only provided the resulting improper integral converges. This only means if you first integrate from $a$ to $x,$ get that answer, and then let $a to -infty,$ that limit exists.



I'm assuming $f(t)$ is defined and continuous (or sufficiently nice) on $(-infty,x).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:20












  • $begingroup$
    @user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
    $endgroup$
    – coffeemath
    Dec 8 '18 at 1:26










  • $begingroup$
    @user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
    $endgroup$
    – eyeballfrog
    Dec 8 '18 at 5:23
















3












$begingroup$

Yes, lower limit can be $-infty$ but only provided the resulting improper integral converges. This only means if you first integrate from $a$ to $x,$ get that answer, and then let $a to -infty,$ that limit exists.



I'm assuming $f(t)$ is defined and continuous (or sufficiently nice) on $(-infty,x).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:20












  • $begingroup$
    @user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
    $endgroup$
    – coffeemath
    Dec 8 '18 at 1:26










  • $begingroup$
    @user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
    $endgroup$
    – eyeballfrog
    Dec 8 '18 at 5:23














3












3








3





$begingroup$

Yes, lower limit can be $-infty$ but only provided the resulting improper integral converges. This only means if you first integrate from $a$ to $x,$ get that answer, and then let $a to -infty,$ that limit exists.



I'm assuming $f(t)$ is defined and continuous (or sufficiently nice) on $(-infty,x).$






share|cite|improve this answer









$endgroup$



Yes, lower limit can be $-infty$ but only provided the resulting improper integral converges. This only means if you first integrate from $a$ to $x,$ get that answer, and then let $a to -infty,$ that limit exists.



I'm assuming $f(t)$ is defined and continuous (or sufficiently nice) on $(-infty,x).$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 0:54









coffeemathcoffeemath

2,7501415




2,7501415












  • $begingroup$
    why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:20












  • $begingroup$
    @user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
    $endgroup$
    – coffeemath
    Dec 8 '18 at 1:26










  • $begingroup$
    @user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
    $endgroup$
    – eyeballfrog
    Dec 8 '18 at 5:23


















  • $begingroup$
    why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:20












  • $begingroup$
    @user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
    $endgroup$
    – coffeemath
    Dec 8 '18 at 1:26










  • $begingroup$
    @user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
    $endgroup$
    – eyeballfrog
    Dec 8 '18 at 5:23
















$begingroup$
why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
$endgroup$
– drerD
Dec 8 '18 at 1:20






$begingroup$
why does the function f(x) needs to be continuous? Is the integral of a non-continuous f(x) non-continuous?
$endgroup$
– drerD
Dec 8 '18 at 1:20














$begingroup$
@user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
$endgroup$
– coffeemath
Dec 8 '18 at 1:26




$begingroup$
@user14042 Integration theory does not require integrand to be continuous. But it gets complicated in more general cases, and at least one needs to know if it's Riemann integral or other type.
$endgroup$
– coffeemath
Dec 8 '18 at 1:26












$begingroup$
@user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
$endgroup$
– eyeballfrog
Dec 8 '18 at 5:23




$begingroup$
@user14042 Consider the case of a function $f(x)$ that is equal to $e^x$ for non integers and equal to $x$ for integers. If you integrate it and take the derivative, you'll get $e^x$, which is a different function from $f$. Continuity is sufficient to ensure this doesn't happen.
$endgroup$
– eyeballfrog
Dec 8 '18 at 5:23











0












$begingroup$

f(x) depeds only on x not on a.



If f(t)=-t then $frac{d}{dx} intlimits^x_a (-t)dt = -frac{x^2}{2}$ and the lower boun of this function is $-infty$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what's lim sup?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:17
















0












$begingroup$

f(x) depeds only on x not on a.



If f(t)=-t then $frac{d}{dx} intlimits^x_a (-t)dt = -frac{x^2}{2}$ and the lower boun of this function is $-infty$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what's lim sup?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:17














0












0








0





$begingroup$

f(x) depeds only on x not on a.



If f(t)=-t then $frac{d}{dx} intlimits^x_a (-t)dt = -frac{x^2}{2}$ and the lower boun of this function is $-infty$.






share|cite|improve this answer











$endgroup$



f(x) depeds only on x not on a.



If f(t)=-t then $frac{d}{dx} intlimits^x_a (-t)dt = -frac{x^2}{2}$ and the lower boun of this function is $-infty$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 1:47

























answered Dec 8 '18 at 1:06









Станчо ПавловСтанчо Павлов

12




12












  • $begingroup$
    what's lim sup?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:17


















  • $begingroup$
    what's lim sup?
    $endgroup$
    – drerD
    Dec 8 '18 at 1:17
















$begingroup$
what's lim sup?
$endgroup$
– drerD
Dec 8 '18 at 1:17




$begingroup$
what's lim sup?
$endgroup$
– drerD
Dec 8 '18 at 1:17


















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Grease: Live!

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A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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