Evaluating a double integral involving a biquadratic with variable coefficients

Question

Please keep in mind that I only have an intermediate level proficiency in Calculus-II . I am relatively new to the concept of double integration, though I have solved some problems in this topic.

I have tried a lot, but have failed in simplifying this integral

$$\int_{u=0}^1 \int_{t=0}^{\infty} \frac{2 \sqrt{115} \cdot (27t^2+11)}{\left[(729+1035u^2)t^4+(594+2070u^2)t^2 + (121+1035u^2)\right]} dt du$$

I know that in general, to deal with integrals of the type

$$\int \frac{ax^2+b}{x^4+px^2+q} dx$$

We divide both numerator and denominator by $x^2$ and try to find suitable constants $c,d$ such that:

$$ a + \frac{b}{x^2} = c \left(1 + \frac{\sqrt{q}}{x^2}\right) + d \left(1 - \frac{\sqrt{q}}{x^2}\right)$$

But after I take $(729+1035u^2)$ common to make the integrand in such a form, I am not able to re-integrate the expression I obtain with respect to $u$. The integrals look terrible and certainly unsolvable by me.

WolframAlpha suggests that the answer of the integral may be $\frac{\pi^2}{9}$

I would like to know of any better methods to solve this double integral. Thanks in advance! Have a nice day!

Answer 1

We can factor the integrand into something nicer and less intimidating.

The integrand is equal to

$$I = \int_0^1 \int_0^{\infty} \frac{2 \sqrt{115} \cdot(27t^2+11)}{(27t^2+11)^2+1035u^2\cdot (t^2+1)^2} dt du$$

Swapping the order of integration, we get

$$ I = \int_{t=0}^{\infty} \int_{u=0}^1 \frac{2 \sqrt{115} \cdot(27t^2+11)}{(27t^2+11)^2+u^2\cdot (3\sqrt{115})^2\cdot (t^2+1)^2} du dt$$

$$ = \int_{t=0}^{\infty} \frac{2\sqrt{115}(27t^2+11)}{1035(t^2+1)^2} \cdot \int_{u=0}^1 \frac{1}{\frac{(27t^2+11)^2}{\left[3 \sqrt{115} \cdot (t^2+1)\right]^2} + u^2} du dt$$

$$ = \int_{t=0}^{\infty} \frac{2\sqrt{115}(27t^2+11)}{1035(t^2+1)^2} \cdot \frac{3\sqrt{115}(t^2+1)}{(27t^2+11)} \cdot \arctan \left(\frac{3\sqrt{115}(t^2+1)}{(27t^2+11)} \right) dt$$

$$ = \frac{2}{3} \cdot \int_0^{\infty} \frac{1}{(t^2+1)} \cdot \arctan \left(\frac{3\sqrt{115}(t^2+1)}{(27t^2+11)} \right) dt$$

Now let us substitute $t = \tan \theta$. The integral the becomes:

$$I = \frac{2}{3} \cdot \int_0^{\frac{\pi}{2}} \arctan \left( \frac{3\sqrt{115} \sec^2\theta}{27\sec^2\theta - 16}\right) d\theta$$

$$ I = \frac{2}{3} \cdot \int_0^{\frac{\pi}{2}} \arctan \left(\frac{3\sqrt{115}}{11+16 \sin^2\theta} \right) d\theta$$

$$ = \frac{2}{3} \cdot \int_0^{\frac{\pi}{2}} \arctan \left(\frac{3\sqrt{115}}{19 - 8 \cos 2\theta} \right) d\theta$$

Substitute $2\theta = x$ and we get:

$$ I = \frac{1}{3} \cdot \int_0^{{\pi}} \arctan \left(\frac{3\sqrt{115}}{19 - 8 \cos x} \right) dx$$

$$ = \frac{1}{3} \cdot \int_0^{\pi} \frac{\pi}{2} dx - \frac{1}{3} \cdot \int_0^{\pi} \arctan \left(\frac{19 - 8 \cos x}{3\sqrt{115}} \right) dx$$

$$I = \frac{\pi^2}{6} - \frac{1}{3} \cdot \int_0^{\pi} \arctan \left(\frac{19 - 8 \cos x}{3\sqrt{115}} \right) dx$$

Note that the second integral is nothing but

$$I_1 = \int_0^{\pi} \arctan \left(\frac{19 - 8 \cos x}{3\sqrt{115}} \right) dx = \frac{\pi^2}{6}$$

Because, observe that

$$4\cdot \left(\frac{19}{3\sqrt{115}}\right)^2 - \left(\frac{8}{3\sqrt{115}}\right)^2 = \frac{4}{3}$$

And for $0<a<1$, $b>0$ and $4a^2-b^2 = \frac{4}{3}$, we have

$$\int_0^{\pi} \arctan \left( a + b\cos x \right) dx = \int_0^{\pi} \arctan \left( a - b\cos x \right) dx = \frac{\pi^2}{6}$$

The proof of which is given by Sangchul Lee, linked here..

Finally, our main integral becomes:

$$I = \frac{\pi^2}{6} - \frac{1}{3} \cdot \frac{\pi^2}{6} = \frac{\pi^2}{6} - \frac{\pi^2}{18} = \frac{\pi^2}{9}$$

Which is the desired result and matches with WolframAlpha.

Answer 2

Integrate w.r.t. $u$ first; the denominator can be rearranged as

$$\left(1035 t^4 + 2070 t^2 + 1035\right) u^2 + \left(729 t^4 + 594 t^2 + 121\right) = 1035 \left(t^2+1\right)^2 u^2 + \left(27t^2+11\right)^2$$

$$\implies \int_0^\infty \left[\int_0^1 \cdots \, du\right] \, dt = \frac23 \int_0^\infty \frac{\arctan \left(\frac{3\sqrt{115}\left(t^2+1\right)}{27t^2+11}\right)}{t^2+1} \, dt \tag1$$

Integrate by parts to "simplify" the $\arctan$ piece; we end up with

$$\frac\pi3 \arctan\frac{\sqrt{115}}9 + \frac{48\sqrt{115}}3 \int_0^\infty \frac{t \arctan t}{441t^4 + 666t^2 + 289} \, dt$$

The latter integral is amenable to complex methods. See the general result derived here. If that's something unfamiliar to you, check out the linked posts therein for non-complex methods. Considering $\dfrac{\pi^2}9=\dfrac23\cdot\dfrac{\pi^2}6$, as Scipio's answer demonstrates, one can recast $(1)$ into a special case of this