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Derivative of a function under an integral sign for inventory control

0

$begingroup$

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

share|cite|improve this question

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

0

$begingroup$

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

share|cite|improve this question

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

share|cite|improve this question

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

$endgroup$

share|cite|improve this question

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

share|cite|improve this question

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

edited Apr 2 at 10:02

Max

1,0021319

edited Apr 2 at 10:02

Max

1,0021319

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32

asked Apr 2 at 9:40

FromTheoryToPracticeFromTheoryToPractice

32