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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 ?

Pinteco
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  • Maybe I misunderstood, but doesn't $\frac{d^n}{dx^n}x^n=n!$? – Noble Mushtak Oct 20 '18 at 23:59
  • Yes, but I will not use this fact to give the example, so I can solve my doubt. – Pinteco Oct 21 '18 at 00:01
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    Even in single-variable calculus, though, integration by parts only uses first derivatives. You are using $n^{\text{th}}$ derivatives, and that doesn't work for integration by parts. – Noble Mushtak Oct 21 '18 at 00:02
  • I think in the post has the answer, the problem are the limits of the integral. https://math.stackexchange.com/questions/1532817/repeated-integration-by-parts-of-a-definite-integral – Pinteco Oct 21 '18 at 00:08
  • It says, "The boundary conditions required are $f^{(k)}(b)g^{(n-k)} (b) = f^{(k)}(a)g^{(n-k)} (a)$ for all $k \in [0:n-1]$," but I don't think your choice of $f$ and $g$ here satisfy those conditions. – Noble Mushtak Oct 21 '18 at 00:10
  • Yes, I didn't know about this requirement, this must be why the result is off. – Pinteco Oct 21 '18 at 00:12

2 Answers2

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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: $$ \int_0^1 \biggl( \text{here you put what you get when integrating the inner integral by parts} \biggr) \, dy . $$

Hans Lundmark
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  • does the integration of parts of only the inner integral also apply to the case of an unbounded support (indefinite integral)? $\int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(x,y) , dx \right) , dy$ – develarist Dec 22 '20 at 08:45
  • @develarist: Yes, provided that you take the same precautions that you would do in single-variable analysis (since, as I wrote, the way it's written, the inner integral is nothing but a single-variable integral with $y$ playing the role of a constant). – Hans Lundmark Dec 22 '20 at 08:57
  • In the case of two variables, how to show the steps in-between of $$\int \int (x + y) e^{x + y} , dx , dy = e^{x + y} (x + y - 2) + c_1 x + c_2 $$ – develarist Dec 22 '20 at 09:00
  • @develarist: What does that have to do with improper integrals? You should ask that as a separate question. – Hans Lundmark Dec 22 '20 at 09:03
  • is it correct to say $$\iint_{[0,1]^2} c(u,v) \ \mathrm{d}u , \mathrm{d}v = \int_0^1 c(u,v) \bigg|_{v=0}^{v=1} \mathrm{d}u $$ as I have done here: https://math.stackexchange.com/questions/3829062/archimedean-clayton-copula-entropy – develarist Jan 04 '21 at 17:10
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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.

Noble Mushtak
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