General form of a double integral

Question

I have difficulty calculating the general form of a double integral:

$\int_0^1\int_0^1\sqrt{x^2+y^2-2mxy}\,\text{d}x\text{d}y$, where m is a constant between -1 and 1.

I've tried some special cases:

m=0:

$$ \int_0^1\int_0^1\sqrt{x^2+y^2}\text{d}x\text{d}y =2\int_0^{\frac{\pi}4}\int_0^{\sec\theta}r^2\text{d}r\text{d}\theta =\frac13\int_0^{\frac{\pi}4}\sec^3\theta\text{d}\theta=\frac13(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)\big|_0^{\frac{\pi}{4}}$$

$$ =\frac{\sqrt2+\ln(\sqrt2+1)}{3}$$

m=1:

$$ \int_0^1\int_0^1|x-y|\text{d}x\text{d}y=\frac13 $$

General cases:

$$ \int_0^1\int_0^1\sqrt{x^2+y^2-2mxy}\text{d}x\text{d}y=\int_0^1\int_0^1\sqrt{(x-y\cos\theta)^2+y^2\sin^2\theta}\text{d}x\text{d}y $$

I've tried to use substitution, but it seemed more complex...

Update:Actually I've tried to continue, here's what I got:

$$ \int\sqrt{(x-y\cos\theta)^2+y^2\sin^2\theta}\text{d}(x-y\cos\theta) $$

Using substitution $x-y\cos\theta=y\sin\theta\tan u, \text{d}(x-y\cos\theta)=y\sin\theta\sec^2 u$

$$=y\sin\theta\int\sec^3 u\text{d}u=\frac12x\sec u-\frac12y\sin\theta\tan\theta+\frac12\ln|\sec u+\frac{x}{y\sin\theta}-\tan\theta|$$....

Update:

Thanks to @Po1ynomial in the comments, I tried to use polar coordinate but got stuck here:

$$2\int_0^{\frac{\pi}4}\int_0^{\sec\theta}r^2\sqrt{1-m\sin2\theta}\text{d}r\text{d}\theta=\frac23\int_0^{\frac{\pi}4}\sec^3\theta\sqrt{1-m\sin2\theta}\text{d}\theta=?...$$

Answer 1

Thanks to @Po1ynomial in the comments.

Using polar coordinates: $$I=\int_0^1\int_0^1\sqrt{x^2+y^2-2mxy}\text{d}x\text{d}y=2\int_0^{\frac{\pi}4}\int_0^{\sec\theta}r\sqrt{1-2m\sin\theta\cos\theta}\text{d}r\text{d}\theta=\frac23\int_0^{\frac{\pi}4}\sec^3\theta\sqrt{1-2m\sin\theta\cos\theta}\text{d}\theta$$

Substitute $t=\tan\theta$: $$I=\frac23\int_0^1\sqrt{1+t^2-2mt}\text{d}t=\frac23\int_0^1\sqrt{(t-m)^2-m^2+1}\text{d}t$$

We first consider:

$$\int\sqrt{(t-m)^2-m^2+1}\text{d}(t-m)$$

Substitute $t-m=\sqrt{1-m^2}\tan u$,$\text{d}(t-m)=\sqrt{1-m^2}\sec^2u$:

$$(1-m^2)\int\sec^3u\text{d}u=\frac{1-m^2}{2}(\sec u\tan u+\ln|\sec u+\tan u|)$$

Notice that $\sec u =\sqrt{1+\frac{(t-m)^2}{1-m^2}}=\sqrt{\frac{t^2-2mt+1}{1-m^2}}$, $\tan u =\frac{t-m}{\sqrt{1-m^2}}$:

As for the case where t=1:

$$\frac{1-m^2}{2}(\sqrt{\frac2{1+m}}\cdot\sqrt{\frac{1-m}{1+m}}+\ln|\sqrt{\frac2{1+m}}+\sqrt{\frac{1-m}{1+m}}|)=\frac{(1-m)^\frac32}{\sqrt2}+\frac{1-m^2}{2}(\ln(\sqrt2+\sqrt{1-m})-\ln\sqrt{1+m})$$

t=0:

$$-\frac{m}{2}+\frac{1-m^2}{2}\ln\frac{\sqrt{1-m}}{\sqrt{1+m}}$$

So:

$$I=\frac{\sqrt2}3(1-m)^{\frac32}+\frac{m}3+\frac13(1-m^2)\ln\frac{\sqrt{2(1-m)}+1-m}{1-m}$$

Answer 2

To solve $$\int_0^1\int_0^1\sqrt{x^2 + y^2 + 2mxy}dxdy$$, begin by using the completing the square method to get $$\int_0^1\int_0^1\sqrt{(x^2 + 2mxy + m^2y^2) - m^2y^2 + y^2}dxdy = \int_0^1\int_0^1\sqrt{(x + my)^2 - m^2y^2 + y^2}dxdy$$ Now, substitute $$u = x + my, x = u - my, du = dx$$ This slightly simplifies the integral and makes a complicated trigonometric substitution possible. Now, make another substitution. This substitution is $$v = tan^-1{\frac{u}{\sqrt{y^2 - m^2y^2}}}, u = tan(v)\sqrt{y^2 - m^2y^2}, du = sec^2\sqrt{y^2 - m^2y^2} dv$$ This substitution may seem random, but it will allow the usage of the formula $$ytan^2(x) + y = ysec^2(x)$$ Now, the integral looks like this: $$\int_0^1\int_0^1\sec^2(v)\sqrt{y^2 - m^2y^2}\sqrt{(y^2 - m^2y^2)tan^2(v) - m^2y^2 + y^2} dv = \int_0^1\sqrt{y^2 - m^2y^2}\sqrt{y^2 - m^2y^2}\int_0^1\sec^3(v)dvdy$$ Finally, the reduction formula can be applied to solve the first integral. By the way, the way everything is written may look weird because I don't really know how something like this should look like. Also, another trigonomic formula is used, $$sec(tan^-1(x)) = \sqrt{x^2 + 1}$$ $$\int_0^1\sqrt{y^2 - m^2y^2}\sqrt{y^2 - m^2y^2}\frac{(sec(v)tan(v))}{2} + \frac{1}{2}\int_0^1sec(v)dvdy = \int_0^1y^2-m^2y^2(\frac{(sec(v)+tan(v)}{2} * \frac{1}{2})ln(tan(v)+sec(v))dy =\int_0^1\frac{(1-m^2)y^2ln(\sqrt{\frac{(x+my)^2}{y^2-m^2y^2} + 1} + \frac{x+my}{\sqrt{y^2-m^2y^2}})}{2} + \frac{(1-m)^2y^2(x+my)\sqrt{\frac{(x+my)^2}{y^2-m^2y^2} + 1}}{2\sqrt{y^2-m^2y^2}}dy$$ Finally, plug 0 and 1 into x to finally complete the first intgral. I will do both integrals separately to save time and not write the massive integral out. The first integral is $$\int_0^1\frac{(1-m^2)y^2ln(\sqrt{\frac{(1+my)^2}{y^2-m^2y^2} + 1} + \frac{1+my}{\sqrt{y^2-m^2y^2}})}{2} + \frac{(1-m)^2y^2(1+my)\sqrt{\frac{(1+my)^2}{y^2-m^2y^2} + 1}}{2\sqrt{y^2-m^2y^2}}dy$$ Start by solving the first term in this equation. Begin by using the formula $$ln(\sqrt{x+1} + x) = sinh^-1(x) *Another update, solving this problem has taken forever, and writing it down takes even longer. I've sadly not had time to complete the problem. I'm planning to finish it over the weekend when I have more time.