calculus - A definite integral with trigonometric functions: $int_{0}^{pi/2} x^{2} sqrt{tan x} sin(2x) , mathrm{d}x$

How could we get a closed form for the following integral:

$$\int_{0}^{\frac{\pi }{2}}{{{x}^{2}}\sqrt{\tan x}\sin \left( 2x
\right) \, \mathrm{d}x} \tag{1}$$

While the antiderivative of $\sqrt{\tan x} \sin(2x)$ can be expressed in terms of elementary functions according to Wolfram Alpha, the antiderivative of $x^{2} \sqrt{\tan x} \sin(2x)$ seeminly cannot be expressed in closed form.

To evaluate $(1)$, would introducing a parameter and differentating under the integral sign be helpful?

Answer

The general method of attack on an integral like this is to recognize that the $x^2$ could be ignored for the time being upon expressing the integral in terms of a suitable parameter which may then be twice differentiated. To wit, after a little manipulatin, I find:

$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = 2 \Im{ \int_{0}^{\frac{\pi}{2}} dx \: x^2 e^{i x} \sqrt{\sin{x} \cos{x}}} $$

So consider the following integral:

$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: e^{i \alpha x} \sqrt{\sin{x} \cos{x}} $$

Note that the integral we seek is

$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = -2 \Im{\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1}} $$

It turns out that there is, in fact, a closed for for $J(\alpha)$:

$$J(\alpha) = -\frac{\left(\frac{1}{16}+\frac{i}{16}\right) \sqrt{\frac{\pi }{2}} \left ( e^{\frac{i \pi a}{2}}-i\right) \Gamma \left(\frac{a-1}{4}\right)}{\Gamma \left ( \frac{a+5}{4} \right )} $$

Plugging this into the above expression, you will find terms including polylogs and harmonic numbers, which I will spare you unless explicitly asked for. But this is how you would get a closed-form expression for your integral.

EDIT

The integral is even nicer when we consider

$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: \sin{\alpha x} \sqrt{\sin{x} \cos{x}} $$

Then

$$J(\alpha) = \frac{\pi ^{3/2} \sin \left(\frac{\pi a}{4}\right)}{8 \Gamma
\left(\frac{5-a}{4}\right) \Gamma \left(\frac{5+a}{4}\right)}$$

and

$$\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-5 \pi ^2+6 \pi (\log (4)-2)+3 (8+(\log (4)-4) \log
(4))\right)}{192 \sqrt{2}}$$

The integral we seek is then

$$\int_0^{\frac{\pi}{2}} dx \: x^2 \sqrt{\tan{x}} \sin{(2 x)} = -2 \left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-24+12 \pi +5 \pi ^2-3 \log ^2(4)+12 \log (4)-6 \pi \log
(4)\right)}{96 \sqrt{2}}$$

It turns out that the numerical value of the latter value is about $1.10577$, which agrees with the numerical approximation of the integral mentioned by @mrf and verified in Mathematica.