difficult transformation and integration, double integral

[dts5044]

use the change of variables x = (u-v)/SQRT(2), y = (u+v)/SQRT(2) to evaluate

[integral from 0 to 1] [integral from 0 to 1] 1/(1-xy) * dxdy

i am given the hint to use the identity (1 - sint)/cost = cost/(1+sint) = tan(pi/4 -t/2)

so i need to prove the identities (have done so...no problem with that so i wont post it), transform the integral into terms of u and v, and then integrate!

the domain in xy is a square in the first quadrant, vertices (0,0) (0,1) (1,0) and (1,1)

let L[sub:a7h940bn]1[/sub:a7h940bn] be the bottom edge of the square

y = 0, 0 <= x <=1

i.e. v = -u, 0 <= u <= 1/SQRT(2)
-----------------------------------------------
let L[sub:a7h940bn]2[/sub:a7h940bn] be the right edge of the square

0 <= y <= 1, x = 1

i.e. v = u - SQRT(2), 1/SQRT(2) <= u <= SQRT(2)
-----------------------------------------------
let L[sub:a7h940bn]3[/sub:a7h940bn] be the top edge of the square

y = 1, 0 <= x <=1

i.e. v = SQRT(2) - u, 1/SQRT(2) <= u <= SQRT(2)
------------------------------------------------
let L[sub:a7h940bn]4[/sub:a7h940bn] be the left edge of the square

0 <= y <= 1, x = 0

i.e. v = u, 0 <= u <= 1/SQRT(2)
-------------------------------------------------

so when all is said and done, the transformation from the domain D relating to xy to the domain T relating to uv is a square in the uv plane with vertices (0,0) (1/SQRT(2), 1/SQRT(2)) ((SQRT(2), 0) and (1/SQRT(2), -1/SQRT(2))

also the Jacobian is 1

please if you can double check this, as I'm not positive I got it right

now, from here I can't figure out how to evaluate the integral! Please help if you can, thanks!!

 

[Unco]

Hi Dts,

Your work so far is good. Notice the transformation you have chosen is a clockwise rotation by pi/4.

Call the originial integral I. Then, after substituting for x and y, using the symmetry of the rotated square and integrating first with respect to v, we have \(\displaystyle I = 4\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{\sqrt{2-u^2}}\arctan{\frac{u}{\sqrt{2-u^2}} du + 4\int_{\frac{1}{\sqrt{2}}}^2 \frac{1}{\sqrt{2-u^2}}\arctan{\frac{\sqrt{2}-u}{\sqrt{2-u^2}} du\),
which is the tricky bit. One approach then, since \(\displaystyle \frac{d}{du}\sin{\frac{u}{\sqrt{2}}}=\frac{1}{\sqrt{2-u^2}}\), is to turn the arctans into arcsins. The first one is obvious (draw a triangle); the latter requires a bit more work. For the latter, let \(\displaystyle \theta = \arctan{\frac{\sqrt{2}-u}{\sqrt{2-u^2}}}\); then from \(\displaystyle \tan^2{\theta}\), one may get \(\displaystyle \cos{2\theta}\) and hence \(\displaystyle \theta = \frac{\pi}{4} - \frac{1}{2}\arcsin{\frac{u}{\sqrt{2}}\).

Upon reflecting, however, it would be quicker just to use the substitution \(\displaystyle u = \sqrt{2}\cos{2\theta}\) for the second integral.

Let's see how you go.

 

[galactus]

We can do this with Jacobians, but a geometric series may work as well.

\(\displaystyle I=\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy\)

Expand \(\displaystyle \frac{1}{1-xy}\) as a geometric series, decompose the summands as products, and integrate:

\(\displaystyle I=\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(xy)^{n}dxdy=\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}x^{n}y^{n}dxdy\)

\(\displaystyle =\sum_{n=0}^{\infty}\left(\int_{0}^{1}x^{n}dx\right)\left(\int_{0}^{1}y^{n}dy\right)=\sum_{n=0}^{\infty}\frac{1}{n+1}\cdot\frac{1}{n+1}\)

\(\displaystyle =\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}=\sum_{n=1}^{\infty}\frac{1}{n^{2}}\)

You know what this famous sum evaluates to?.

If you must to do the Jacobian thing:

Let \(\displaystyle u=v=\frac{y-x}{2}\) be the new coordinates.

The domain of integration is a square of side length \(\displaystyle \frac{\sqrt{2}}{2}\), which we get from the old domain first by rotating it by 45 degrees

and then shrinking it a factor of \(\displaystyle \sqrt{2}\). Subbing of \(\displaystyle x=u-v, \;\ y=u+v\) gives us:

\(\displaystyle \frac{1}{1-xy}=\frac{1}{1-u^{2}+v^{2}}\)

To transform the integral, we have to replace dxdy by 2dudv to compensate for the fact that the coordinate transform reduces the area by a factor of

2.

We get:

\(\displaystyle I=4\int_{0}^{\frac{1}{2}}\left(\int_{0}^{u}\frac{dv}{1-u^{2}+v^{2}}\right)du+4\int_{\frac{1}{2}}^{1}\left(\int_{0}^{1-u}\frac{dv}{1-u^{2}+v^{2}}\right)du\)

Now, use \(\displaystyle \int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}tan^{-1}(\frac{x}{a})\).

Can you finish?. Perhaps learn a little LaTex. It is difficult to read what you posted.