How can I find the following integral:

$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx $$

My thoughts:

Can we possibly convert this to spherical or use change of variables?


according to the shape of the area of integration and the shape of the function that is under integral, the easiest answer is to define variables $u=x+y$ and $v=x-y$then we have: $$\frac{1}{J}=\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1&1\\1&-1\end{vmatrix}\Rightarrow |J|=\frac{1}{2}$$ and the borders of the area of integration are $u=-v,u=1,v=0$ just draw the shape of the problem to understand this part.then:
$$\int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx=\int^{0}_{-1}\int^1_{-v} \frac{1}{2}u^9v^9\,du\,dv=\frac{-1}{400}$$

Edit: to answer the question asked in the comments:
why did I use $\frac{1}{|J|}$?
consider the analog situation for a 1-D integral where we have to use change of variables method:
we have $u=x^2\Rightarrow du=2x\,dx$
meaning that we have $u$ as a function of $x$ so we can calculate $du=\frac{du}{dx}dx$ but in order to substitute $dx$ in the original integral we write $du=2x\,dx\Rightarrow dx=\frac{1}{2x}du$ which means that we need $dx=\frac{dx}{du}du$ to substitute in the original integral so
In a 1-D integral we have $u$ as a function of $x$ so we can calculate $\frac{du}{dx}$ but to substitute in the original integral we need $\frac{dx}{du}$
here is the same situation:
to substitute in the original integral, we need:
But most of the time we have $u$ and $v$ as a function of $x$ and $y$:
so we can compute $$\frac{1}{|J|}=\frac{\partial(u,v)}{\partial(x,y)}$$

  • $\begingroup$ Does the definition of your new variables have something to do with an affine transformation? $\endgroup$ – Khallil Jul 27 '15 at 20:37
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    $\begingroup$ @Khallil as you know we can use the method change of variables in 1-D integrals to calculate them easier just like that we can use change of variables in a 2-D or 3-D integral. and the jacobi above defines the relationship between old variables and new ones. change of variables is just mapping from a space to another space just like the affine transformation is. $\endgroup$ – Sepideh Abadpour Jul 27 '15 at 21:06
  • $\begingroup$ May I ask why you chose to define your variables in such a way? $\endgroup$ – Khallil Jul 27 '15 at 21:14
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    $\begingroup$ @Khallil because I want to find a way to calculate the above integral easier. if you draw the borders of the original integral in the question you will see that it's a triangle sides $x+y=1,x-y=0,x=0$ also you have the terms $x-y,x+y$ in the function so you will deduce that choosing the variables this way will make both the function and the borders easier. I should say that choosing the new variables is just a trick you will learn as much as you solve problem.practice makes perfect $\endgroup$ – Sepideh Abadpour Jul 27 '15 at 21:27
  • $\begingroup$ Thanks for the help, @sepideh! $\endgroup$ – Khallil Jul 27 '15 at 21:28

Hint: $$(x+y)^9(x-y)^9=((x+y)(x-y))^9=(x^2-y^2)^9$$

  • $\begingroup$ Okay, so we can introduce an 'r here by saying that $(-(x^2+y^2))^9=(-r^2)^9=-r^{18}$. Right? $\endgroup$ – Nadia Marson Jul 27 '15 at 19:53
  • $\begingroup$ Okay I got that much. Now what do I do next? $\endgroup$ – Nadia Marson Jul 27 '15 at 20:03
  • $\begingroup$ Actually, the expression is $(x^2-y^2)^9$. Now you can use binomial expansion of $(x^2-y^2)^9$ so that all the terms are separated in powers of $x$ & $y$ then integrating all, first w.r.t. $y$ then w.r.t. $x$ It can be done but the procedure/expansion will be lengthy. $\endgroup$ – Harish Chandra Rajpoot Jul 27 '15 at 20:17
  • $\begingroup$ It doesn't seem efficient or intuitive. Just a hammering of algebra. $\endgroup$ – Khallil Jul 27 '15 at 20:39

How about a change of variable like $u=x+y, v=x-y$?

The Jacobian is $-\frac 12$, and the area of integration is the triangle bounded by the lines $x=y, x+y=1, x=0$

This translates as: $v$ varies from $v=0$ to $ v=u$ for the inner integral, and $u=0$ to $u=1$ for the outer integral.

Therefore we evaluate $$\int_{u=0}^{u=1}\int_{v=0}^{v=u}u^9v^9(-\frac 12)dvdu$$

The final answer is $-\frac{1}{400}$

  • $\begingroup$ Yes. Actually I was wanting something in this direction. Can you please show me this step-by-step if you don't mind? $\endgroup$ – Nadia Marson Jul 27 '15 at 19:56

\begin{align} u & = x+y \\ v & = x-y \end{align} $$ du\,dv = \left|\frac{\partial (u,v)}{\partial(x,y)}\right|\,dx\,dy = 2\,dx\,dy $$

\begin{align} & \int^{1/2}_0 \int^{1-x}_x (x+y)^9(x-y)^9 \, dy \, dx \\[10pt] = {} & \int_0^1 \left( \int_{u-1}^0 u^9 v^9 2\,dv \right) \,du \\[10pt] = {} & \int_0^1 u^9 \frac{(-(u-1)^{10})} 5 \, du \end{align}


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