What is the Derivative of an Integral with a Variable Limit?

[cappadonza]

hey I'm having problems trying to find the derivative of the function

i'm not sure how the separate the x out of the integral

i want to find the derivative of F(x) where
[tex] F(x) = \int_{+\infty}^{x^2} e^{-xt^2} dt [/tex]

thanks

 

[rasmhop]

Let
[tex]P(x) = \int e^{-xt^2} \textrm{ d}t[/tex]
(with arbitrary constant)
Then you probably know how to find P'(x), but the hard part is finding
[tex]\frac{dP(x^2)}{dx}[/tex]
Let s be the squaring function (i.e. let s(x) = x^2), then what you want to find is:
[tex]\frac{d(P\circ s)(x)}{dx}[[/tex]
which you can find using the chain-rule, and then you just do:
[tex]F'(x) = \frac{dP(x^2)}{dx} - \left(\lim_{t \to +\infty} P(t)\right)'[/tex]

 

[cappadonza]

Thanks i did do it the way you suggested but what confused me is that there is an e^-x

under the integral. I was trying to find a way to it in a form so that only t was under the integral.
For example when
i had
[tex] F(x) = \int e^{-(x +t)} dt [/tex]
i got it into the following form

[tex] F(x) = e^-x \int e^-t dt [/tex]
from here i use the product rule to find the derivative of F(x).

I thought i would have had to do something similar to move
[tex] e^{-x} [/tex]
outside the integral. or doesn't it matter in this case since i can treat as a constant in the power

thanks

 

[D H]

hey I'm having problems trying to find the derivative of the function

i'm not sure how the separate the x out of the integral

i want to find the derivative of F(x) where
[tex] F(x) = \int_{+\infty}^{x^2} e^{-xt^2} dt [/tex]

thanks

The general rule for taking the derivative of a definite integral is the Leibiz integral rule:

[tex]\frac d{dx}\int_{a(x)}^{b(x)} f(x,t) dt =
\frac{db(x)}{dx} f(x,b(x)) \,-\, \frac{da(x)}{dx} f(x,a(x)) \,+\,
\int_{a(x)}^{b(x)} \frac{\partial f(x,t)}{\partial x} dt[/tex]

Applying that rule to this specific problem is not all that tough. (Is this homework?)BTW, did you really mean to use +infinity for the lower limit on your integral?

 

[cappadonza]

Thanks
no this not a homework question, its been many years since my first degree, I'm basically going through basic calculus notes to get ready for my post graduate course in mathematics in finance.
i did mean to use +infinity, since the exponential is to the - ve power, this shouldn't be a problem since this will tend to 0 as t-> infinity. does this make sense or am i missing something

 

[D H]

Since this is not homework, applying the Leibniz integral rule to this particular integral yields

[tex]
\frac {dF(x)}{dx} = \frac d{dx}\int_{\infty}^{x^2} e^{-xt^2}\,dt
= 2x e^{-x^3} - \int_{\infty}^{x^2} t^2e^{-xt^2}\,dt \qquad\qqaud (1)
[/tex]

That integral on the right-hand side can be re-expressed in terms of the original integral. The easiest way is to integrate the original integral, [itex]F(x)=\int \exp(-xt^2)\,dt[/itex], by parts:

[tex]\aligned
u &= e^{-xt^2} & du &= -2xte^{-xt^2}\,dt \\
dv &= dt & v &= t\endaligned[/tex]

Integrating by parts,

[tex]
\int e^{-xt^2}\,dt = te^{-xt^2} + 2x\int t^2 e^{-xt^2}\,dt
[/tex]

or

[tex]
\int t^2 e^{-xt^2}\,dt = \frac 1{2x}\left(\int e^{-xt^2}\,dt - te^{-xt^2}\right)
[/tex]

Applying this in (1) yields (assuming I didn't make a dumb mistake somewhere),

[tex]\aligned
\frac {dF(x)}{dx}
&= 2x e^{-x^3} -
\frac 1{2x}
\left(
\int_{\infty}^{x^2} e^{-xt^2}\,dt -
\left(\left. t e^{-xt^2}\right|_{t=\infty}^{x^2}\right)
\right)
&= \frac 5 2\,x e^{-x^3} - \frac {F(x)}{2x}
[/tex]

 

[cappadonza]

Thanks makes a lot of sense