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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Aardman Animations

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Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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