Challenging Integral ${\int_{0}^{\frac{\pi}{2}} \left(\frac{x\ln(\sin(x))}{\sqrt{\tan(x)}}\right) dx}$

Question

I came across this integral last week during my Calc II self-study sessions:

$\qquad \boxed{\int_{0}^{\frac{\pi}{2}} \left(\frac{x\ln(\sin(x))}{\sqrt{\tan(x)}}\right) dx}$

I had taken most of my leisure time to research recognizable parts of this definite integral. I even came across videos and MSE posts explaining some of these integrals and the different ways you can solve them (i.e., differentiation under the integral sign):

$1.\quad \boxed{\int_{0}^{\frac{\pi}{2}} \left( \sqrt{\tan(x)} \right) dx}$

$2.\quad \boxed{\int_{0}^{\frac{\pi}{2}} \left( \frac{1}{\sqrt{\tan(x)}} \right) dx}$

$3.\quad \boxed{\int_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ dx}$

I've done about $15$ related integrals if I can recall correctly [this post was supposed to be scheduled for others to review $2$ weeks ago, but was abandoned and never posted to MSE because I had to prepare and practice my upcoming May AS Level exams]

An MSE post about computing the third integral contains a solution using The Beta Function, which I fail to understand, and would appreciate it if someone explained it to me.

I have tried visualizing what $\frac{x\ln(\sin(x))}{\sqrt{\tan(x)}}$ would look like in a graphing calculator between $0$ and $\frac{\pi}{2}$, the function sticks out into the $(+x,-y)$ quadrant, and evaluating its integral within the boundaries yields a negative area.

Firstly, I tried solving this integral using a substitution method like IBP and U-sub, but got nowhere. I have also attempted to solve this integral by differentiation under the integral sign, but ended up with the following unfinished result:

• $I'(t)=\displaystyle\int_{0}^{\frac{\pi}{2}} x^2\cot(tx)\sqrt{\tan(x)} \ dx$

I have also watched a video that talks about trigonometric functions hiding in factorials, but I'm not really sure how I can represent a function like this with Gamma and Digamma functions..

Is there a closed-form solution for this integral? If so, then what methods did you use? Please do provide a full working as well so I can teach it to myself and understand it.

Answer 1

Start by substituting $x\to\dfrac\pi2-x$ to utilize Quanto's result, denoted by $Q$ below, then deal with the remaining integral using the beta function:

$$\begin{align*} I &= \int_0^\tfrac\pi2 x \sqrt{\cot x} \log(\sin x) \, dx \\ &= \frac\pi2 \underbrace{\int_0^\tfrac\pi2 \sqrt{\tan x} \log(\cos x) \, dx}_J - Q & x\to\frac\pi2-x \\[2ex] J &= \frac14 \int_0^1 \frac{\log y}{y^{3/4} (1-y)^{1/4}} \, dy & y=\sqrt{\cos x} \\ &= \frac14 \frac\partial{\partial a} \int_0^1 y^{a-\frac34} (1-y)^{-\frac14} \, dy \,\bigg|_{a=0} \\ &= \frac14 \Gamma\left(\frac34\right) \frac\partial{\partial a} \operatorname{Beta}\left(\frac34,a+\frac14\right) \bigg|_{a=0} \\ &= -\frac\pi{4\sqrt2} \left(\pi+6\log2\right) \end{align*}$$

Combining everything points to an overall closed form of

$$I = \boxed{\frac\pi{4\sqrt2} \left(3\log^22 - 2\pi\log2 - \frac{\pi^2}6 + 4G\right)}$$

where $G$ is Catalan's constant.

Answer 2

Applying substitution $ u = \tan x $, $ du = \sec^2 x \, dx = (1 + u^2) \, dx $, $ dx = \frac{du}{1 + u^2} $, $ x = \arctan u $, $ \sin x = \frac{u}{\sqrt{1 + u^2}} $, limits $ u: 0 \to \infty $.

$$ \frac{x \ln(\sin x)}{\sqrt{\tan x}} \, dx = \frac{\arctan u \ln\left(\frac{u}{\sqrt{1 + u^2}}\right)}{\sqrt{u}} \cdot \frac{du}{1 + u^2} $$

Using $ \ln\left(\frac{u}{\sqrt{1 + u^2}}\right) = \ln u - \frac{1}{2} \ln(1 + u^2) $:

$$ I = \int_0^\infty \frac{\arctan u \left( \ln u - \frac{1}{2} \ln(1 + u^2) \right)}{\sqrt{u} (1 + u^2)} \, du = I_1 - \frac{1}{2} I_2 $$

Evaluating $ I_1 = \int_0^\infty \frac{\arctan u \ln u}{\sqrt{u} (1 + u^2)} \, du $.

Using parameterized integral $ J(\alpha) = \int_0^\infty \frac{\arctan u \, u^\alpha}{(1 + u^2)} \, du $, so $ I_1 = \left. \frac{\partial J}{\partial \alpha} \right|_{\alpha = -1/2} $.

Substituting $ u = \tan \theta $:

$$ J(\alpha) = \int_0^{\pi/2} \theta \tan^\alpha \theta \, d\theta = \frac{1}{2} B\left(\frac{\alpha + 1}{2}, \frac{1 - \alpha}{2}\right) \left[ \psi\left(\frac{\alpha + 1}{2}\right) - \psi(1) \right] $$

$$ = \frac{\pi}{2 \sin\left(\pi \frac{\alpha + 1}{2}\right)} \left[ \psi\left(\frac{\alpha + 1}{2}\right) + \gamma \right] $$

Differentiating:

$$ \frac{\partial J}{\partial \alpha} = \frac{\pi}{2} \left[ -\frac{\pi}{2} \frac{\cos\left(\pi \frac{\alpha + 1}{2}\right)}{\sin^2\left(\pi \frac{\alpha + 1}{2}\right)} \left( \psi\left(\frac{\alpha + 1}{2}\right) + \gamma \right) + \frac{1}{\sin\left(\pi \frac{\alpha + 1}{2}\right)} \cdot \frac{1}{2} \psi'\left(\frac{\alpha + 1}{2}\right) \right] $$

At $ \alpha = -1/2 $:

$$ I_1 = \pi \sqrt{2} \ln 2 + \frac{3 \pi^3 \sqrt{2}}{8} + 2 \pi \sqrt{2} G $$

Evaluating $ I_2 = \int_0^\infty \frac{\arctan u \ln(1 + u^2)}{\sqrt{u} (1 + u^2)} \, du $.

Substituting $ v = u^2 $:

$$ I_2 = \frac{1}{2} \int_0^\infty \frac{\arctan \sqrt{v} \ln(1 + v)}{v^{1/2} (1 + v)} \, dv $$

Substituting $ w = \sqrt{v} $:

$$ I_2 = \int_0^\infty \frac{\arctan w \ln(1 + w^2)}{1 + w^2} \, dw $$

Substituting $ w = \tan \theta $:

$$ I_2 = \int_0^{\pi/2} \theta \ln(\sec^2 \theta) \, d\theta = 2 \int_0^{\pi/2} \theta \ln \sec \theta \, d\theta = -2 \int_0^{\pi/2} \theta \ln \cos \theta \, d\theta $$

Using $ \int_0^{\pi/2} \theta \ln \cos \theta \, d\theta = -\frac{\pi^2}{8} \ln 2 - \frac{7 \pi}{8} G $:

$$ I_2 = \frac{\pi^2}{4} \ln 2 + \frac{7 \pi}{4} G $$

Combining:

$$ I = I_1 - \frac{1}{2} I_2 = \left( \pi \sqrt{2} \ln 2 + \frac{3 \pi^3 \sqrt{2}}{8} + 2 \pi \sqrt{2} G \right) - \frac{1}{2} \left( \frac{\pi^2}{4} \ln 2 + \frac{7 \pi}{4} G \right) $$

$$ = \pi \sqrt{2} \ln 2 - \frac{\pi^2}{8} \ln 2 + \frac{3 \pi^3 \sqrt{2}}{8} + 2 \pi \sqrt{2} G - \frac{7 \pi}{8} G $$

Therefore, the answer is:

$$ \boxed{\pi \sqrt{2} \ln 2 - \frac{\pi^2}{8} \ln 2 + \frac{3 \pi^3 \sqrt{2}}{8} + 2 \pi \sqrt{2} G - \frac{7 \pi}{8} G} $$

where G is Catalan's constant

This doesn't match numerically, but this is my best effort. I hope it can help somebody figure out this integral

Answer 3

Let us first solve an easier integral related to your question and then understand how we can proceed.

Let $J=\displaystyle\int_0^\frac\pi2\frac{\ln{\sin x}}{\sqrt{\tan x}} \,dx$.

So, let us take a function $J(\alpha)$ st.
$J(\alpha)=\int_{0}^{\frac{\pi}{2}} (\sin x)^{\alpha} (\cos x)^{\frac{1}{2}} \,dx$

clearly $J=J'\left(\frac{-1}2\right)$.

Now, let us find $J(\alpha)$.

$$J(\alpha)=\frac{1}{2}B\left(\frac{\alpha+1}{2},\frac{3}{4}\right)$$

$$J(\alpha)=\frac{1}{2}\frac{\Gamma(\frac{\alpha+1}{2})\Gamma(\frac{3}{4})}{\Gamma(\frac{2\alpha+5}{4})}$$

Now, differentiate with respect to $\alpha$ and put $\alpha=-\frac{1}{2}$:

$$J=\frac{1}{4}\left(\Gamma'({\frac{1}{4}})-\Gamma({\frac{1}{4}})\Gamma'(1)\right)$$

Since, $\Gamma'(1)=-\gamma$ and $\psi(z)=\frac{\Gamma'({z})}{\Gamma(z)}$

$$\boxed{J=\frac14\Gamma\left(\frac34\right)\left(\psi\left(\frac14\right)+\gamma)\right)}$$

Now the problem with your integral is that it has an extra $x$ in the numerator and using similar techniques cause the function with parameter alpha say $I(\alpha)$ to have divergent terms so we cannot proceed with this method. But nonetheless we can solve the question without the $x$ factor.

I tried with $(\tan x)^\alpha$, but it got too complicated; but I think it can be done.

Answer 4

Not a solution, yet too long to be a comment:

Simply taking $x\mapsto\frac{\pi}{2}-x$, your integral equals to $$ \int_{0}^{\pi/2} \frac{x\ln(\sin x)}{\sqrt{\tan x}}\,\mathrm{d}x = \frac{\pi}{2} \int_{0}^{\pi/2} \sqrt{\tan x}\ln(\cos x)\,\mathrm{d}x - \int_{0}^{\pi/2} x\sqrt{\tan x}\ln(\cos x)\,\mathrm{d}x $$

First integral actually has a close-form due to the special case of Beta function at quarters, even without the knowledge of Beta function we still can let $u=\sqrt{\tan x}$ $$ \begin{aligned} \int_{0}^{\pi/2} \sqrt{\tan x}\ln(\cos x)\,\mathrm{d}x & = -\int_{0}^{\infty} \frac{u^2\ln(1+u^4)}{1+u^4}\,\mathrm{d}u \\ & = -\int_{0}^{1} \frac{u^2\ln(1+u^4)}{1+u^4}\,\mathrm{d}u - \int_{1}^{\infty} \frac{u^2\ln(1+u^4)}{1+u^4}\,\mathrm{d}u \\ (u\mapsto1/u)\; & = -\int_{0}^{1} \frac{u^2\ln(1+u^4)}{1+u^4}\,\mathrm{d}u - \int_{0}^{1} \frac{\ln\bigr(1+\frac1{u^4}\bigl)}{1+u^4}\,\mathrm{d}u \\ & = -\int_{0}^{1} \frac{(1+u^2)\ln\bigr(u^2+\frac1{u^2}\bigl)}{1+u^4}\,\mathrm{d}u + 2\int_{0}^{1} \frac{(1-u^2)\ln u}{1+u^4}\,\mathrm{d}u \\ \left(v=\frac1{u}-u\right) & = -\int_{0}^{\infty} \frac{\ln(2+v^2)}{2+v^2}\,\mathrm{d}v + 2\int_{0}^{1} \frac{(1-u^2)\ln u}{1+u^4}\,\mathrm{d}u \end{aligned} $$ where $$ \int_{0}^{\infty} \frac{\ln(2+v^2)}{2+v^2}\,\mathrm{d}v = \frac{3\pi\ln2}{2\sqrt2},\quad \int_{0}^{1} \frac{(1-u^2)\ln u}{1+u^4}\,\mathrm{d}u = -\frac{\pi^2}{8\sqrt{2}} $$ may not be trivial, yet much easier to solve (the right one related to the derivative of digamma function). Nevertheless the second part $$ \int_{0}^{\pi/2} x\sqrt{\tan x}\ln(\cos x)\,\mathrm{d}x $$ is way more harder to crack.