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Derivative of a function under an integral sign for inventory control
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Moment Generating Function for distance from center in a circle, uniform PDF.Integral of a Kernel functionDifferentiation through the integral sign (Lebesgue integration)Problem with differentiation under integral signHow to find the derivative of an integral where both, the limit and the integrand, are functions of x?Differentiation under the integral sign for an electrostatic fieldObtaining cdf from a pdf containing absolute value and sign functionA differentiation under the integral signTime Derivative Under Integral SignFind constant,PDF,mean value from CDF
$begingroup$
I have found the derivative of a function within an integral by numerical computation. However, I can't figure out how to do it mathematically correct.
The complete function is the integral, going from '$r$' to plus infinity. $r$ is a constant and $> 0$.
The integrand is $(x-r)cdot f(x)$, where x is a random variable, normally distributed and $f(.)$ is the probability density function (pdf) of the normal distribution. After this expression, '$dx$' follows of course. I don't know how to write it in LaTeX.
For the derivative of this with respect to $r$, I found: $F(r)-1$, where $F(.)$ is the cumulative density function (cdf). If we change the 'barriers' on the integral to '$r$' on top of the integral sign and zero at the bottom of the integral sign, I find as a derivative $1-F(r)$.
Can anybody help to derive this mathematically correct and tell me how I should do it?
Thanks!
Steven
integration discrete-mathematics derivatives partial-derivative
$endgroup$
add a comment |
$begingroup$
I have found the derivative of a function within an integral by numerical computation. However, I can't figure out how to do it mathematically correct.
The complete function is the integral, going from '$r$' to plus infinity. $r$ is a constant and $> 0$.
The integrand is $(x-r)cdot f(x)$, where x is a random variable, normally distributed and $f(.)$ is the probability density function (pdf) of the normal distribution. After this expression, '$dx$' follows of course. I don't know how to write it in LaTeX.
For the derivative of this with respect to $r$, I found: $F(r)-1$, where $F(.)$ is the cumulative density function (cdf). If we change the 'barriers' on the integral to '$r$' on top of the integral sign and zero at the bottom of the integral sign, I find as a derivative $1-F(r)$.
Can anybody help to derive this mathematically correct and tell me how I should do it?
Thanks!
Steven
integration discrete-mathematics derivatives partial-derivative
$endgroup$
$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25
add a comment |
$begingroup$
I have found the derivative of a function within an integral by numerical computation. However, I can't figure out how to do it mathematically correct.
The complete function is the integral, going from '$r$' to plus infinity. $r$ is a constant and $> 0$.
The integrand is $(x-r)cdot f(x)$, where x is a random variable, normally distributed and $f(.)$ is the probability density function (pdf) of the normal distribution. After this expression, '$dx$' follows of course. I don't know how to write it in LaTeX.
For the derivative of this with respect to $r$, I found: $F(r)-1$, where $F(.)$ is the cumulative density function (cdf). If we change the 'barriers' on the integral to '$r$' on top of the integral sign and zero at the bottom of the integral sign, I find as a derivative $1-F(r)$.
Can anybody help to derive this mathematically correct and tell me how I should do it?
Thanks!
Steven
integration discrete-mathematics derivatives partial-derivative
$endgroup$
I have found the derivative of a function within an integral by numerical computation. However, I can't figure out how to do it mathematically correct.
The complete function is the integral, going from '$r$' to plus infinity. $r$ is a constant and $> 0$.
The integrand is $(x-r)cdot f(x)$, where x is a random variable, normally distributed and $f(.)$ is the probability density function (pdf) of the normal distribution. After this expression, '$dx$' follows of course. I don't know how to write it in LaTeX.
For the derivative of this with respect to $r$, I found: $F(r)-1$, where $F(.)$ is the cumulative density function (cdf). If we change the 'barriers' on the integral to '$r$' on top of the integral sign and zero at the bottom of the integral sign, I find as a derivative $1-F(r)$.
Can anybody help to derive this mathematically correct and tell me how I should do it?
Thanks!
Steven
integration discrete-mathematics derivatives partial-derivative
integration discrete-mathematics derivatives partial-derivative
edited Apr 2 at 10:02
Max
1,0021319
1,0021319
asked Apr 2 at 9:40
FromTheoryToPracticeFromTheoryToPractice
32
32
$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25
add a comment |
$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25
$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25
add a comment |
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$begingroup$
You can use the Leibniz integral rule
$endgroup$
– callculus
Apr 2 at 12:32
$begingroup$
thanks! It was exactly the Leibniz rule that I needed! Stupid of me not to think about that.
$endgroup$
– FromTheoryToPractice
Apr 4 at 7:25