# Derivative of an integral

1. Dec 12, 2009

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

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

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

thanks

#### Attached Files:

• ###### integral.gif
File size:
759 bytes
Views:
39
2. Dec 12, 2009

### rasmhop

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

3. Dec 12, 2009

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
$$F(x) = \int e^{-(x +t)} dt$$
i got it into the following form

$$F(x) = e^-x \int e^-t dt$$
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
$$e^{-x}$$
outside the integral. or doesnt it matter in this case since i can treat as a constant in the power

thanks

Last edited: Dec 12, 2009
4. Dec 12, 2009

### D H

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

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

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?

5. Dec 12, 2009

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 shouldnt be a problem since this will tend to 0 as t-> infinity. does this make sense or am i missing something

6. Dec 12, 2009

### D H

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

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

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, $F(x)=\int \exp(-xt^2)\,dt$, by parts:

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

Integrating by parts,

$$\int e^{-xt^2}\,dt = te^{-xt^2} + 2x\int t^2 e^{-xt^2}\,dt$$

or

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

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

\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}

7. Dec 13, 2009