3

Statement of the problem

Let $F(x,y)=\iint_{R_{xy}} e^{(u-1)}e^{(v^2-v)}dudv$ where $R_{xy}$ is the region $[0,x]\times[0,y]$. Calculate $\nabla F(1,1)$

The attempt at a solution

I know that $\nabla F(1,1)=(\dfrac{\partial F(1,1)}{\partial x},\dfrac{\partial F(1,1)}{\partial y})$.

I am going to calculate the first partial derivative since the other one can be calculated in a similar way.

$\dfrac{\partial F (x,y)}{\partial x}=\dfrac{\partial}{\partial x} \int_0^y\int_0^x e^{(u-1)}e^{(v^2-v)}dudv$. Now, I don't know if the following step is legitimate:

$\dfrac{\partial}{\partial x} \int_0^y\int_0^x e^{(u-1)}e^{(v^2-v)}dudv=\int_0^y [\dfrac{\partial}{\partial x} \int_0^x e^{(u-1)}e^{(v^2-v)}du]dv$

If that last step was correct, I would like to know how to justify it.

Now, by the fundamental theorem of calculus I know that

$\dfrac{\partial}{\partial x} \int_0^x e^{(u-1)}e^{(v^2-v)}du=e^{(x-1)}e^{(v^2-v)}$

Using this I get

$\dfrac{\partial F (x,y)}{\partial x}=\int_0^y e^{(x-1)}e^{(v^2-v)}dv=e^{(x-1)}\int_0^y e^{(v^2-v)}dv$

Here I got stuck, I've tried to calculate this integral but I couldn't.

Have I done something wrong up to now and maybe that's why I am having trouble with this integral? I would appreciate some help.

user100106
  • 3,573
  • 7
  • 41
  • 64

1 Answers1

1

The step you want to justify is actually a pretty subtle thing, known as differentiation under the integral sign. If you're new to calculations like this, then you probably aren't expected to justify it, but this answer of Qiaochu Yuan tells you some conditions under which it's valid.

You applied the Fundamental Theorem of Calculus correctly. The integral you got stuck on has no elementary antiderivative, and doesn't really simplify to something in terms of simple constants when you plug in $(1,1)$ as you have to do to get $\frac{\partial F(1,1)}{\partial x}$. So after you plug in $(1,1)$ and simplify the $e^{x-1}$ part, there's nothing more you can really do with that piece symbolically.

The $y$ partial should work out a bit more nicely.

Community
  • 1
Mark S.
  • 25,893
  • I never heard of "differentiation under integral sign" (Leibniz integral rule) until now. Thanks for your answer! – user100106 Feb 23 '14 at 03:48