# A hard definite integral with trignometric

How could we get a closed form for this one? $$\displaystyle\int_{0}^{\frac{\pi }{2}}{{{x}^{2}}\sqrt{\tan x}\sin \left( 2x \right)\text{d}x}$$

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Why do you expect there to be one? A quick numeric computation and a look at the inverse symbolic calculator gives no match for the number (1.1057733282580962321535756282112900456642421892605) – mrf Jan 23 at 10:38
@mrf the answer is in terms of pai and logs. I dont know what substitution or other methods are involved. – Ryan Jan 23 at 10:47
Using @rlgordonma's method, the final result can be written in terms of elementary functions and is given by $\frac{\pi \left(5 \pi ^2-6 \pi (\log (4)-2)-3 (8+(\log (4)-4) \log (4))\right)}{96 \sqrt{2}}$. – Fabian Jan 23 at 11:21
@Fabian: I get precisely what you get. I take it we have Mathematica in common? And I can simplify what I have presented and present the actual result if it is wanted. – Ron Gordon Jan 23 at 11:52
@rigordonma I am really interested how to get these value :) – Ryan Jan 23 at 13:20
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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

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

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.

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Very nice! That was surprising, at least to me. – mrf Jan 23 at 13:03
@mrf: Thanks. That means a lot given what I've seen you do here. – Ron Gordon Jan 23 at 20:18