Does integrating by parts in two variables works the same as one variable?

Question

I will give a example that I tried to integrate by parts in $x$:

\begin{align} \int_0^1 \int_0^1 ye^x \frac{d^n}{dx^n} x^n\:\: dxdy \end{align} Using the rule of integration by parts:

$f= ye^x \implies f^{(n)} = ye^x$

$g'= \frac{d^n}{dx^n}x^n \implies g = x^n$

I found the $n$ th derivative and integrated $n$ times.

\begin{align} &\int_0^1 \int_0^1 ye^x \frac{d^n}{dx^n} x^n\:\: dxdy = \\ &fg\vert_0^1 - \int_0^1 \int_0^1 f'g \: \:dxdy=\\ & ye^x x \vert_0^1 - \int_0^1 \int_0^1 ye^x x^n \:\:dxdy=\\ & y e - \int_0^1 \int_0^1 ye^x x^n \:\:dxdy\\ \end{align}

The left side is a function of $y$ and the right side is a function of $n$. Since $n$ a natural number we can evaluate the right side but why is the left side a function of $y$? Shouldn't be the integral a function of $n$?

Is this how we integrate by parts in two variables ?

Answer 1

This is not how you do integration-by-parts in multivariable calculus. There are a lot of possible theorems for this, but I will show you this one:

$$\int \int_{\Omega} \frac{\partial u}{\partial x}vdxdy=\int_{\Gamma}uv\hat n_xd\Gamma-\int \int_{\Omega}u\frac{\partial v}{\partial x}dxdy$$

Here, $\Omega$ is the area of integration (in this case, $0 \leq x,y \leq 1$) and $\Gamma$ is the boundary of integration (in this case, it is a square of the following pattern: $x=1$ from $y=0$ to $y=1$, $y=1$ from $x=1$ to $x=0$, $x=0$ from $y=1$ to $y=0$, and $y=0$ from $x=0$ to $x=1$). Also, $\hat n_x$ is the x-component of the outward unit normal to the boundary $\Gamma$, which would be $1$ when on $x=1$, $-1$ when on $x=0$, and $0$ otherwise. However, this integration by parts introduces line integrals in order to calculate a rather simple double integral, so I don't think it will help you here. Nevertheless, it is a pretty interesting theorem, so make of it what you will.

Answer 2

You don't really have a double integral here, since it's already written as an iterated integral $$ \int_0^1 \left( \int_0^1 f(x,y) \, dx \right) \, dy . $$ Here both integrals are in fact standard one-variable integrals (where in the inner integral $y$ is treated like a constant), so you can use all the rules from single-variable calculus. For example, you can do integration by parts, but if you want to do that on the inner integral, you must do it on the inner integral only: $$ \int_0^1 \biggl( \text{here you put what you get when integrating the inner integral by parts} \biggr) \, dy . $$