Lemma : $\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ -ky } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } =\frac{{ (-1) }^{ k }}{\pi} \int _{ k\pi }^{ \infty }{ \frac { sin(x) }{ x } dx }$ , for $k$ is a whole number.

Proof : We being with noting that :

$\displaystyle \frac { 1 }{ w } \int _{ 0 }^{ \infty }{ { e }^{ -st }sinwtdt } =\frac { 1 }{ { s }^{ 2 }+{ w }^{ 2 } }$

So we can rewrite the integral on the left side as :

$\displaystyle I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ { e }^{ -ky }dy } \int _{ 0 }^{ \infty }{ { e }^{ -ty }sin\pi tdt } =\frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ sin\pi t\int _{ 0 }^{ \infty }{ { e }^{ -(k+t)y }dy } dt }$

$\displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ 0 }^{ \infty }{ \frac { sin\pi t }{ t+k } dt }$

$\displaystyle \Rightarrow I = \frac { 1 }{ \pi } \int _{ k }^{ \infty }{ \frac { sin(\pi t-\pi k) }{ t } dt }$

$\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k }^{ \infty }{ \frac { sin(\pi t) }{ t } dt }$

$\displaystyle \Rightarrow I = { (-1) }^{ k }\int _{ k\pi }^{ \infty }{ \frac { sin(t) }{ t } dt }$

Hence proved

Solution :

We start with $x={e}^{-y}$ to get out integral as :

$\displaystyle I = \int _{ -\infty }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$

Break in into two parts :

$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ -\infty }^{ 0 }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$

In the second part put $y=-x$ to get :

$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -x }\ln { (1+{ e }^{ x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dy }$

Manipulating it we have :

$\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { e }^{ y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { e }^{ -y }\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$

$\displaystyle \Rightarrow I= \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ y }+{ e }^{ -y })\ln { (1+{ e }^{ -y }) } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy } +\int _{ 0 }^{ \infty }{ \frac { { ye }^{ -y } }{ { y }^{ 2 }+{ \pi }^{ 2 } } dy }$

We write $I = J+K$

$\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x })\ln { (1+{ e }^{ -x }) } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx$

Wrting $\ln(1+{e}^{-x})$ in it's taylor series we have :

$\displaystyle J = \int _{ 0 }^{ \infty }{ \frac { ({ e }^{ x }+{ e }^{ -x }) }{ { x }^{ 2 }+{ \pi }^{ 2 } } \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ -rx } } } dx$

We interchange the integral and sum :

$\displaystyle J = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } (\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r-1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx+\int _{ 0 }^{ \infty }{ \frac { { e }^{ -(r+1)x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } } dx } )$

Using the lemma we have :

$\displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ (r-1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } +\int _{ (r+1)\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } } dx) }$

Now it is worthy noting that :

$\displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx }$ , for all positive $r$, what if it was applied for $r=0$, then the left integral is $\frac{\pi}{2}$, while the write one is 0, so for $r=0$, we can write it as :

$\displaystyle \int _{ r\pi }^{ \infty }{ \frac { \sin { (x) } }{ x } dx } =\int _{ 1 }^{ \infty }{ \frac { sin(r\pi x) }{ x } dx } +\frac { \pi }{ 2 }$, for $r=0$, having said that we proceed further :

$\displaystyle J = \frac { 1 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { \sin { (2(r-1)\pi x) } +\sin { (2(r+1)\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }$

$\displaystyle \Rightarrow J = \frac { 2 }{ \pi } \sum _{ r=1 }^{ \infty }{ \frac { 1 }{ r } (\int _{ \frac{1}{2} }^{ \infty }{ \frac { cos(2\pi x)\sin { (2r\pi x) } }{ x } dx } ) } +\frac { 1 }{ 2 }$

Again changing sum and integral we have :

$\displaystyle \Rightarrow J = \frac { 2 }{ \pi } \int _{ \frac { 1 }{ 2 } }^{ \infty }{ \cos { (2\pi x) } \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ rx } } dx } +\frac { 1 }{ 2 }$

It is a well known result that :

$\displaystyle \sum _{ r=1 }^{ \infty }{ \frac { \sin { (2r\pi x) } }{ r\pi } } =\left\lfloor x \right\rfloor +\frac { 1 }{ 2 }-x$

Using this we have :

$\displaystyle J = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 }$

Leaving this let's evaluate $K$ first :

Using the same techniques as used in proving the lemma, we can show that :

$\displaystyle K = \int _{ 0 }^{ \infty }{ \frac { x{ e }^{ -x } }{ { x }^{ 2 }+{ \pi }^{ 2 } } dx } =-\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx }$

Now $\displaystyle I = J+K = 2\int _{ \frac { 1 }{ 2 } }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } (-x+\left\lfloor x \right\rfloor +\frac { 1 }{ 2 } )dx } +\frac { 1 }{ 2 } -\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } =\frac { 1 }{ 2 } +2\int _{ 1/2 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$

$\displaystyle I = \frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx } +2\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \cos { (2\pi x) } dx } =\frac { 1 }{ 2 } +2\int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$

Now we will be evaluating :

$\displaystyle M = \int _{ 1 }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } \left\{ x \right\} dx }$

It can be written as :

$\displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } (x-n)dx } }$

$\displaystyle \Rightarrow M = -\sum _{ n=1 }^{ \infty }{ n\int _{ n }^{ n+1 }{ \frac { \cos { (2\pi x) } }{ x } dx } }$

$M$ can also be written as :

$\displaystyle M =-\sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \cos { (2\pi x) } }{ x } dx } }$

This manipulation can easily be justified with the help of properties of double summations :

Using integration by parts we have :

$\displaystyle M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } } -\sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi n) } }{ 2\pi n } }$

$\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ \infty }{ \frac { \sin { (2\pi x) } }{ 2\pi { x }^{ 2 } } dx } }$

$\displaystyle \Rightarrow M = \sum _{ n=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } dx } }$

Again changing sum and integral we have :

$\displaystyle \Rightarrow M = \quad \int _{ 1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { \sin { (2\pi nx) } }{ 2\pi { nx }^{ 2 } } } dx }$

Using the result we have used above we have :

$\displaystyle M = \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$

$\displaystyle \Rightarrow M = \quad \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { 1 }{ 2{ x }^{ 2 } } -\frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$

For solving $\displaystyle H = \int _{ 1 }^{ \infty }{ \frac { \left\{ x \right\} }{ { x }^{ 2 } } dx }$

write it as a sum :

$\displaystyle H = \sum _{ n=1 }^{ \infty }{ \int _{ n }^{ n+1 }{ \frac { 1 }{ x } -\frac { n }{ { x }^{ 2 } } dx } }$

$\displaystyle \Rightarrow H = \sum _{ n=1 }^{ \infty }{ \ln { (n+1) } -\ln { (n) } -\frac { 1 }{ n+1 } }$

$\displaystyle H = \lim _{ n\rightarrow \infty }{ 1+\ln { (n+1)-{ H }_{ n+1 } } } =1-\gamma$

Hence $\displaystyle M = \frac { \gamma }{ 2 } -\frac { 1 }{ 4 }$

Finally we got :

$\Large \boxed { I=\gamma }$

$\text{Solution to Problem 26}$

Firstly some common things

$2 \sin ^{2} x= 1-\cos (2x)$

$\large{\pi I_{n} (z)=\displaystyle \int^{\pi}_{0} e^{z \cos \theta} \cos (n \theta) d \theta}$

where $I_{n} (z)$ is modified Bessel function of first kind

Also $\large{I_{n}(z) =e^{\frac{-in \pi}{2}} J_{n} (i z)}$ where $i^2=-1$ and $J_{n} (z)$ is Bessel function of first kind.

In this question we have $n=0$

Now in the integral

$\large{\displaystyle \int^{\pi}_{0} e^{2 \cos x} \left(\frac{1- \cos (2 \sin x)}{2}\right) dx}$

On separating

$\large{= \frac{\pi}{2} I_{0} (2) -\frac{1}{2}\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}$

Let $\large{L=\displaystyle \int^{\pi}_{0} e^{2 \cos x} \cos (2 \sin x) dx}$

$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 \cos x}e^{2i \sin x} dx \right)}$

$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2 (\cos x+i \sin x)} dx \right)}$

$\large{L=\Re \left(\displaystyle \int^{\pi}_{0} e^{2e^{ix}} dx \right)}$

$\large{L=\Re \left(\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{(2e^{ix})^{k}}{k!}dx \right)}$

$\large{L=\displaystyle \int^{\pi}_{0}\displaystyle \sum_{k=0}^{\infty} \frac{2^{k} \cos (kx)}{k!}dx}$

Notice that $\large{\displaystyle \int^{\pi}_{0} \cos (kx) dx=0 \quad \quad \forall k \geq 1 \\ \text{k is an integer}}$

Thus only $k=0$ case is considered and hence $L=\pi$

On combining we have

$\large{ \frac{\pi}{2} \left(I_{0}(2)-1 \right)}$

$\large{ \frac{\pi}{2} \left(J_{0}(2i)-1 \right)}$

Substitute x=2t So the equation becomes $\large{\displaystyle \int^{1}_{0} \ { t }^{ -2/3 }{ (1-t) }^{ -1/3 } } dt$ This comes out as $\large{\displaystyle \Gamma (\frac { 1 }{ 3 } )\Gamma (\frac { 2 }{ 3 } )}$ which is equal to $\large{\displaystyle\frac { 2\pi }{ \sqrt { 3 }} }$

Solution to Problem 28:

The following result is directly used in the solution:$\displaystyle \int_0^{\infty} \frac{ \cos (mx)}{(a^2+x^2)^2} dx = \frac{\pi}{4a^3} e^{-am} ( 1 + am)$.
This can be obtained easily by differentiating the integral w.r.t. $a$ evaluated in solution of problem 15 of this contest. Consider,
$\displaystyle I(b) = \int_0^{\infty} \frac{x \sin(bx) }{ (1+ x^2)^3} dx$. The required integral is $\dfrac{\partial^4}{\partial b^4} I(b)$. Integrating by parts, we get,
$\displaystyle I(b) = - \frac{\sin(bx)}{4(1+x^2)^2} \bigg|_0^{\infty} + \frac{b}{4} \int_0^{\infty} \frac{ \cos (bx)}{(1+x^2)^2} dx = \frac{\pi}{16} e^{-b} ( b+ b^2)$

Differentiate four times w.r.t $b$ and put $b = 1$ to get the value of the integral as $\dfrac{\pi}{8e}$.

Solution to Problem 29

$\large{\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x}dx - \displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2 \sin \left( \frac{\pi x}{2}\right)} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\sin \left( \frac{\pi(1- x)}{2}\right) \right)}{1-x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

In the second integral $x \rightarrow -x$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{0}_{-1} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{1+x}dx-\displaystyle \int^{1}_{0} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{1}_{-1} \frac{\ln \left(\cos \left( \frac{\pi x}{2}\right) \right)}{x+1}dx}$

Now in the second integral $x+1 \rightarrow x$

$\large{=\displaystyle \int^{1}_{0} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{2}_{0} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx}$

$\large{= \displaystyle \lim_{a \to 0} \left( \displaystyle \int^{1}_{a} \frac{\ln \left(\frac{\sin (\pi x)}{2} \right)}{x}dx-\displaystyle \int^{2}_{a} \frac{\ln \left(\sin \left( \frac{\pi x}{2}\right) \right)}{x}dx \right)}$

Again in the second integral $\large{\frac{x}{2} \rightarrow x}$

$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx+\displaystyle \int^{1}_{a} \frac{\ln (\sin (\pi x))}{x} dx-\displaystyle \int^{1}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}$

$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\sin (\pi x))}{x} dx \right)}$

We know $\sin \theta \approx \theta$ when $\theta$ approaches zero.

$\large{= \displaystyle \lim_{a \to 0} \left( -\ln 2 \displaystyle \int^{1}_{a} \frac{1}{x} dx -\displaystyle \int^{a}_{\frac{a}{2}} \frac{\ln (\pi x)}{x} dx \right)}$

$\large{= \displaystyle \lim_{a \to 0} \left( \ln 2 \ln a- \ln \pi \ln 2 -\frac{1}{2} \left((\ln a)^2-(\ln \frac{a}{2})^2 \right) \right)}$

using $a^2-b^2=(a-b)(a+b)$ and on simplifying the terms containing $a$ get cancelled and we finally have

$\Large{\boxed{\frac{1}{2} (\ln 2)^2-(\ln \pi) (\ln 2)}}$

Lemma : $\int_{0}^{1} t^{a} \ln(t) dt = -\dfrac{1}{(a+1)^2}$

This lemma is already proved in many problems in this contest.

$\int_{0}^{\pi/8} \ln(\tan(2x))dx =1/2 \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt$

I used the substitution $t=\tan(2x)$.

$\begin{aligned} \int_{0}^{1} \dfrac{\ln(t)}{1+t^2} dt &= \int_{0}^{1} \sum_{i=0}^{\infty}(-1)^{i} t^{2i} \ln(t)dt \\ &= \sum_{i=0}^{\infty}(-1)^{i} \int_{0}^{1} t^{2i} \ln(t) dt \\ &= -\sum_{i=0}^{\infty} (-1)^{i} \dfrac{1}{(2i+1)^2} \\ &= - \beta(2) \\ &= -G \end{aligned}$

Therefore,

$\int_{0}^{\pi/8} \ln(\tan(2x))dx = \dfrac{-G}{2}$

Not sure if this is considered but...

$y=\ln \left(\tan \left(2x\right)\right)$

$x=\frac{\tan ^{-1}\left(e^y\right)}{2}$

$\int _{ 0 }^{ \frac { \pi }{ 8 } }{ \ln { \left( \tan \left( 2x \right) \right) } } dx=\int _{ -\infty }^{ 0 }{ \frac { \tan ^{ -1 } \left( e^{ y } \right) }{ 2 } } dy=-\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy$

A known identity of the catalan constant is $G=\int _{ 0 }^{ \infty }{ \tan ^{ -1 } \left( e^{ -y } \right) } dy$

Hence the integral is $-\frac{G}{2}$

The identity above can be deduced through the Taylor Series of $\tan^{-1}(x)$

Lemma:

Consider the Gamma Function $\Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1} dt$

Take the substitution, $t=pu^n$.

$\Gamma(s) = n p^{s} \int_{0}^{\infty} e^{-pu^n} u^{ns-1} du$

Taking $p= a+ ib$ and comparing imaginary parts we get,

$\int_{0}^{\infty} e^{-au^n} u^{ns-1} \sin \left(bu^n \right) du = \dfrac{\Gamma(s)}{n |p|^s} \sin(\alpha s )$

where $p=a+ib$ and $\alpha$ is the argument of $p$.

1st Integral:

$\int_{0}^{\infty} \sin \left(x^2 \right) dx$

Comparing this with our lemma, we get $a=0$, $b=1$, $n=2$, $s=1/2$. Which gives us that $|p| = 1$ and $\alpha = \pi /2$.

On substitution we get the integral evaluated to be $I_{1} = \dfrac{\sqrt{\pi}}{2 \sqrt{2}}$.

2nd Integral:

$\int_{0}^{\infty} \dfrac{\sin (x^2)}{x} dx$

Take the substitution, $x = t^{1/2}$. We get,

$\dfrac{1}{2} \int_{0}^{\infty} \dfrac{\sin(t)}{t} dt$

This is well known Integral (Dirichlet integral): $\int_{0}^{\infty} \dfrac{\sin(x)}{x} dx = \dfrac{\pi}{2}$.

So, this integral evaluates to be $I_{2} = \dfrac{\pi}{4}$.

3rd Integral:

This can evaluated in similar way as we obtained for first integral. And this is evaluated to be $I_{3} = \dfrac{\sqrt{\pi}}{\sqrt{2}}$.

So, the overall integral $I = I_{1} + I_{2} - I_{3} =\large{ \boxed{ \dfrac{\pi}{4} - \dfrac{\sqrt{\pi}}{2 \sqrt{2}}}}$

$\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \\ = x \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) \Biggl|_{0}^{\pi /2} - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot (x) dx \right) dx$

$\begin{aligned} \int \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx &= \int \left(\dfrac{\ln(t) - 2}{t \sqrt{t}} \right) dt \\ &=2 \int \dfrac{\ln t d(\sqrt{t} ) - \sqrt{t} d(\ln(t))}{(\sqrt{t}) ^2} \\ &= -2\dfrac{\ln(t)}{\sqrt{t}}\\ &= -2 \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} \end{aligned}$

So,

$\begin{aligned} x \left(\int\left(\dfrac{\ln(\sin (x)) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) \Biggl|_{0}^{\pi /2} &= \dfrac{-2x\ln(\sin(x))}{\sqrt{\sin(x)}} \Biggl|_{0}^{\infty}\\ &= 0 - \lim_{a \rightarrow 0} \dfrac{-2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= \lim_{a \rightarrow 0} \dfrac{2a \ln(\sin(a))}{\sqrt{\sin(a)}}\\ &= 0 \end{aligned}$

Above limit is evaluated using L'Hospital rule.

So,

$\int_{0}^{\pi /2} x \left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx = - \int_{0}^{\pi /2} \left(\int\left(\dfrac{\ln(\sin x) - 2}{\sqrt{\sin x}} \right) \cot x dx \right) dx = 2 \int_{0}^{\pi /2} \dfrac{\ln(\sin(x))}{\sqrt{\sin(x)}} dx$

Let $I(a) = \int_{0}^{\pi /2} \sin^{2a-1} (x) dx$. So, our required integral is $I'(1/4)$. By taking $t = \sin ^2 (x)$, the integral transforms to $I(a) = \dfrac{1}{2} \int_{0}^{1} t^{a-1} (1-t)^{-1/2} dt = \dfrac{1}{2} B(a, 1/2) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)}$

Differentiate it w.r.t $a$,

$I'(a) = \dfrac{1}{2} \dfrac{\Gamma(a) \Gamma(1/2)}{\Gamma(a+1/2)} \left( \psi(a) - \psi(a+1/2) \right)$

$I'(1/4) = \dfrac{1}{2} \dfrac{\Gamma(1/4) \Gamma(1/2)}{\Gamma(3/4)} \left( \psi(1/4) - \psi(3/4) \right)$

Using Euler's reflection formula. We get

$\Gamma(1/4) \Gamma(3/4) = \dfrac{\pi}{\sin(\pi/4)} = \sqrt{2} \pi \implies \Gamma(3/4) = \dfrac{\sqrt{2} \pi}{\Gamma(1/4)}$

Also, $\psi(1/4) - \psi(3/4) = \pi \cot(3 \pi /4) = - \pi$.

Therefore the required integral evaluates to

$-\sqrt{\dfrac{ \pi}{8}} \left( \Gamma \left(\dfrac{1}{4} \right) \right) ^2$

Using basic concepts in integration:

$\quad\displaystyle\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\int_0^1x^n\frac{1-x^n}{1-x}\mathrm dx=\int_0^1x^n\sum_{k=0}^{n-1}x^k\ \mathrm dx=\sum_{k=0}^{n-1}\int_0^1x^{n+k}\ \mathrm dx=\sum_{k=0}^{n-1}\frac1{n+k+1}=\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}$

Then:

$\displaystyle\lim_{n\to\infty}\int_0^1\frac{x^n-x^{2n}}{1-x}\mathrm dx=\lim_{n\to\infty}\frac1n\sum_{k=1}^{n}\frac1{1+\frac kn}=\int_0^1\frac1{1+x}\mathrm dx=[\ln(1+x)]_0^1=\ln2$

Substituting $\displaystyle \sum _{ k=0 }^{ \infty }{ x^{ k } } =\frac { 1 }{ 1-x }$

$\int _{ 0 }^{ 1 }{ \sum _{ k=0 }^{ \infty } x^{ k+n }-\sum _{ k=0 }^{ \infty } x^{ k+2n } } dx=\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+n } -\sum _{ k=1 }^{ \infty } \frac { 1 }{ k+2n }$

$={ H }_{ 2n }-{ H }_{ n }=\ln(2)$

The last step is from the approximation $H_n=\gamma+\ln(n)$ as $n\to \infty$