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 \(t=\tan(2x)\).

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

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 \(I = I_{1} + I_{2} - I_{3} =\large{ \boxed{ \dfrac{\pi}{4} - \dfrac{\sqrt{\pi}}{2 \sqrt{2}}}}\)

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

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

So,

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

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 \(n\to \infty\)

\[\int_{0}^{\pi/2} \cot \left(\dfrac{\theta }{2} \right) \sqrt{\cos \theta} \ln \left( \cos \theta \right) d\theta = \int_{0}^{1} \dfrac{\sqrt{t} \ln(t) }{1-t} dt\]

In the above integral I took \(\cos \theta = t\) and I used the identity \(\cot \dfrac{\theta}{2} = \sqrt{\dfrac{1+\cos \theta}{1- \cos \theta}}\).

Let \(I(a) = \int_{0}^{1} \dfrac{t^a}{1-t} dt\). So, we have to find \(I'(1/2)\).

\[\begin{align*} I(a) &= \int_{0}^{1} \dfrac{t^a}{1-t} dt \\ &= \int_{0}^{1} t^a \sum_{n=0}^{\infty} t^n dt \\ &= \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+a} dt \\ &= \sum_{n=0}^{\infty} \dfrac{1}{n+a+1} \end{align*}\]

Differentiate it on both sides w.r.t \(a\). which gives,

\[\dfrac{d}{da} \sum_{n=0}^{\infty} \dfrac{1}{n+a+1} = - \sum_{n=1}^{\infty} \dfrac{1}{(n+a)^2}\]

Take \(a=1/2\) to evaluate the required integral.

\[\sum_{n=1}^{\infty} \dfrac{1}{(n+1/2)^2} = 4\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)^2}\]

But we know that

\[\begin{align*} \dfrac{\pi^2}{6} &= 1 + \dfrac{1}{2^2 } + \dfrac{1}{3^2} + \ldots \\ &= 1 + \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots + \dfrac{1}{4} \left( 1+ \dfrac{1}{2^2 } + \dfrac{1}{3^2} + \ldots \right) \end{align*} \\ \implies 1+ \dfrac{1}{3^2} + \dfrac{1}{5^2} + \ldots = \dfrac{\pi^2}{8} \\ \implies \sum_{1}^{\infty} \dfrac{1}{(2n+1)^2} = \dfrac{\pi^2}{8} - 1 \]

So, \[\int_{0}^{\pi /2} \cot \left(\dfrac{\theta }{2} \right) \sqrt{\cos \theta} \ln \left( \cos \theta \right) d\theta = -4\sum_{n=1}^{\infty} \dfrac{1}{(2n+1)^2} = \boxed{\dfrac{8-\pi^2}{2}}\]

I first apply the fact that \( \cot(x/2) = \frac{1+\cos x}{\sin x} \).

\(\quad\displaystyle\int^{\frac{\pi}{2}}_{0} \cot \left(\frac{\theta}{2} \right) \sqrt{\cos \theta} \ln (\cos \theta) d \theta =\displaystyle\int^{\frac{\pi}{2}}_{0} \frac{1+\cos\theta}{\sin\theta} \sqrt{\cos \theta} \ln (\cos \theta) d \theta =\displaystyle\int^{0}_{1} \frac{1+u}{\sqrt{1-u^2}} \sqrt{u} \ln (u) \frac{-d u}{\sqrt{1-u^2}}\)

\(=\displaystyle\int^{1}_{0} \frac{1+u}{1-u^2} \sqrt{u} \ln (u) d u\)

\(=\displaystyle\int^{1}_{0} \frac{\sqrt{u} \ln (u)}{1-u} d u\)

\(=\displaystyle\int^{1}_{0} \frac{\sqrt{t^2} \ln (t^2)}{1-t^2} 2tdt\)

\(=\displaystyle\int^{1}_{0} \frac{4 t^2 \ln (t)}{1-t^2} dt\)

\(=\displaystyle-\int^{1}_{0} \ln(t) d\left(4t+2\ln(1-t)-2\ln(1+t)\right)\)

\(=- 0 + \displaystyle\int^{1}_{0} \left(4+\frac{2\ln(1-t)}t-\frac{2\ln(1+t)}t\right) dt\)

\(= 4 + \displaystyle\int^{1}_{0} \left( \frac{2\ln(1-t)}t-\frac{2\ln(1+t)}t \right) dt\)

\(= 4 + 2\displaystyle\int^{1}_{0} \frac1t \sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-t)^n-t^n}n dt\)

\(= 4 - 2\displaystyle\int^{1}_{0} \frac1t \sum_{n=1}^{\infty}\frac{2t^{2n-1}}{2n-1} dt\)

\(= 4 - 2\displaystyle\int^{1}_{0} \sum_{n=1}^{\infty}\frac{2t^{2n-2}}{2n-1} dt\)

\(= 4 - 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}= 4 - 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2} + 4\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = 4 - \frac{4\pi^2}6 + \frac{\pi^2}6 =\boxed{ 4 - \frac{\pi^2}2}.\)

Begin with the substitution \(y=\dfrac{x}{2}\) we have :

\(\displaystyle I = 2\int _{ 0 }^{ 1 }{ \frac { \ln { (\frac { 1 }{ 8 } { (\frac { 1 }{ y } -y) }^{ 3 }+1) } }{ (1+{ y }^{ 2 })\ln { (\frac { 1 }{ 2 } (\frac { 1 }{ y } -y)) } } } dy \)

Again we have the substitution \( \dfrac { 1 }{ 2 } (\dfrac { 1 }{ y } -y)=x\) we have :

\( \displaystyle I =\int _{ 0 }^{ \infty }{ \frac { \ln { (1+{ x }^{ 3 }) } }{ (1+{ x }^{ 2 })\ln { (x) } } } dx \)

Honestly I don't believe this integral should converge as it blows up at x=1, but still I will proceed.

From problem 35 we have :

\( \displaystyle I = \frac { 3\pi }{ 4 } +\ln{(2)}\int _{ 0 }^{ \infty }{ \frac { dx }{ ln(x)(1+{ x }^{ 2 }) } } \)

Now the integral can be shown to be equal to 0( although the function to be integrated blows up to infinity)

hence answer is \( I = \dfrac{3\pi}{4} \).

I'm in.

Hellolo sir, could I be part of this discussion? i'm also pretty interested in Integrals and series as well

Put \(\dfrac{1}{1+x}=y\) to get :

\( \displaystyle I = -\int _{ 0 }^{ 1 }{ { y }^{ (-1/4) }{ (1-y) }^{ -3/4 }\ln { (y) } dy } \)

Use the definition of beta function we have :

\( \displaystyle \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }dy } = \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } \)

Differentiating it with respect to \(m\) we have :

\( \displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (\psi (m)-\psi (m+n)) = \int _{ 0 }^{ 1 }{ { y }^{ m-1 }{ (1-y) }^{ n-1 }\ln { (y) } dy } \)

Put \(\displaystyle m=\dfrac{3}{4} , n=\dfrac{1}{4}\) to get :

\( \displaystyle I = \frac { \Gamma (3/4)\Gamma (1/4) }{ \Gamma (1) } (\psi (1)-\psi (3/4)) \)

Using reflection formula for the gamma function we have :

\( \displaystyle \Gamma(1/4)\Gamma(1-1/4) = \dfrac{\pi}{\sin(\dfrac{\pi}{4})}=\sqrt{2}\pi \)

We will use that \( \displaystyle \psi(x+1) + \gamma = \int _{ 0 }^{ 1 }{ \frac { 1-{ t }^{ x } }{ 1-t } dt } \)

\( \displaystyle \Rightarrow \psi(x+1)-\psi(y+1) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ y }-{ t }^{ x } }{ 1-t } dt } \)

\( \displaystyle \Rightarrow \psi(1)-\psi(3/4) = \int _{ 0 }^{ 1 }{ \frac { { t }^{ -1/4 }-1 }{ 1-t } dt }=J \)

For evaluting \(J\) we will put \(t={x}^{4} \), to get :

\(\displaystyle J = 4\int _{ 0 }^{ 1 }{ \frac { 1-x }{ 1-{ x }^{ 4 } } { x }^{ 2 }dx } \)

\( \displaystyle J = 4\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ (1+{ x }^{ 2 })(1+x) } dx } \)

\(\displaystyle J =2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+x } } +\int _{ 0 }^{ 1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } dx } -2\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } } \)

\( \displaystyle J = 3\ln { (2) }-\dfrac{\pi}{2} \)

\( \displaystyle \Rightarrow I = 3\sqrt { 2 } \pi \ln { (2) } -\frac { { \pi }^{ 2 } }{ \sqrt { 2 } } \)

Alternative method:

\[\dfrac{\ln(1+x)}{1+x} = - \sum_{n=1}^{\infty} H_{n}(-x)^{n}\]

Now use RMT. That's it. \(\ddot \smile\)

(fixed)

Here's the synopsis of the proof:

• I'm going to use the fact that \(\displaystyle \sum_{n=1}^\infty \dfrac{H_n^2}{(n+1)^2} = \dfrac{11}{360}\pi^4\). This value can be proven by seeing Cody's or robjohn's answer. (The proof of this is very long so I've omitted the proof)

• And use Parseval's Theorem to evaluate the integral above by first noting that the integrand is an even function.

Recall that:

\[ \displaystyle f(x)=\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { x }^{ n } \quad \Rightarrow \quad f({ e }^{ ix })=\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { e }^{ ixn } \]

Now according to Fourier series,

\[F(x)=\frac { 1 }{ 2 } { a }_{ 0 }+\sum _{ n=1 }^{ \infty }{ { b }_{ n } } { \sin { (nx) } }+\sum _{ n=1 }^{ \infty }{ { a }_{ n } } { \cos { (nx) } }\]

Parseval's Theorem then states that

\[\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ { F(x) }^{ 2 }dx } =\frac { 1 }{ 2 } { a }_{ 0 }^{ 2 }+\sum _{ n=1 }^{ \infty }{ { a }_{ n }^{ 2 }+{ b }_{ n }^{ 2 } } \]

Substituting \(F(x)=\Im { (f({ e }^{ ix })) }\), \({ a }_{ n }=0\), \(b_n=\frac{H_n}{n+1}\) gives

\[\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ \Im({ f(e }^{ ix } } ))^2dx=\sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 } } \]

Applying these facts above, we get

\[\Im (f(e^{ ix }))=\frac { 1 }{ 2 } \left( x-\pi \right) \ln { \left( 2\sin \left( \frac { 1 }{ 2 } x \right) \right) } ,0\le x\le \pi \]

\[\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ \frac { 1 }{ 4 } { (x-\pi ) }^{ 2 }\ln { ^{ 2 } } \left( 2\sin \left( \frac { 1 }{ 2 } x \right) \right) dx } =\frac { 1 }{ 2\pi } \int _{ 0 }^{ \pi }{ { x }^{ 2 }\ln { ^{ 2 } } \left( 2\cos \left( \frac { 1 }{ 2 } x \right) \right) dx } =\sum _{ n=1 }^{ \infty }{ { b }_{ n }^{ 2 } } =\frac { 11 }{ 360 } { \pi }^{ 4 }\]

\[\ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) = \ln \left(e^{ix/2} + e^{-ix/2} \right) = \ln \left( 1 + e^{ix} \right) - i\dfrac{x}{2}\]

So,

\[\begin{align*} \int_{0}^{\pi / 2} \ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) dx &= \int_{0}^{\pi / 2} \ln \left( 1 + e^{ix} \right) dx - i \int_{0}^{\pi / 2}\dfrac{x}{2} dx \\ &= \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx - i \dfrac{\pi^2}{16} \end{align*}\]

We shall evaluate the first integral

\[\begin{align*} \int_{0}^{\pi /2} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} e^{irx}}{r} dx &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \int_{0}^{\pi /2} e^{irx} dx \\ &= \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1}}{r} \left[ \dfrac{e^{irx}}{ir} \right|_{0}^{\pi/2}\\ &= \dfrac{1}{i} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} \end{align*}\]

Now,

\[ e^{ ir\pi /2 }-1=\quad \begin{cases} i-1 & \quad \quad r=4k+1 \\ -2 & \quad \quad r=4k+2 \\ -i-1 & \quad \quad r=4k+3 \\ 0 & \quad \quad r=4k+4 \end{cases}\]

\[\begin{align*}\sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+2}(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+3}(-2)}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+4}(-i-1)}{(4k+3)^{2}} + \sum_{k=0}^{\infty} \dfrac{(-1)^{4k+5}(0)}{(4k+4)^{2}} \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} + 0 \\ &= \sum_{k=0}^{\infty} \dfrac{(i-1)}{(4k+1)^{2}} + \sum_{k=0}^{\infty} \dfrac{2}{(4k+2)^{2}} + \sum_{k=0}^{\infty} \dfrac{-i-1}{(4k+3)^{2}} \end{align*}\]

\[\sum_{k=0}^{\infty} \dfrac{1}{(4k+1)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+2)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) \\ \sum_{k=0}^{\infty} \dfrac{1}{(4k+3)^2} = \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)\]

So,

\[\begin{align*} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= (i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{4} \right) + 2 \dfrac{1}{16} \zeta \left(2 , \dfrac{1}{2} \right) + (-i-1) \dfrac{1}{16} \zeta \left(2 , \dfrac{3}{4} \right)\\ &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right) \end{align*}\]

1:
\[\zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) = 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} - \dfrac{1}{(4k+3)^2}\right) = 16 \sum_{k=0}^{\infty} \dfrac{(-1)^{k}}{(2k+1)^2} = 16G\]

2:

\[\begin{align*} \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) &= 16 \sum_{k=0}^{\infty} \left(\dfrac{1}{(4k+1)^2} + \dfrac{1}{(4k+3)^2}\right) \\ &= 16 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2}\\ &=16 \dfrac{\pi^2}{8}\\ &= 2 \pi^2 \end{align*}\]

3:

\[\zeta \left(2 , \dfrac{1}{2} \right) = 4 \sum_{k=0}^{\infty} \dfrac{1}{(2k+1)^2} = 4 \dfrac{\pi^2}{8} = \dfrac{\pi^2}{2}\]

So,

\[\begin{align*} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} &= \dfrac{1}{16} \left( i \left( \zeta \left(2 , \dfrac{1}{4} \right) - \zeta \left(2 , \dfrac{3}{4} \right) \right) - \left( \zeta \left(2 , \dfrac{1}{4} \right) + \zeta \left(2 , \dfrac{3}{4} \right) \right) + 2 \zeta \left(2 , \dfrac{1}{2} \right) \right)\\ &= \dfrac{1}{16} \left( i (16G) - (2\pi^2) + 2 \dfrac{\pi^2}{2} \right)\\ &= iG - \dfrac{\pi^2}{16} \end{align*} \]

\[\begin{align*} \int_{0}^{\pi / 2} \ln \left( 2\cos \left(\dfrac{x}{2} \right) \right) dx &= \dfrac{1}{i} \sum_{r=1}^{\infty} \dfrac{(-1)^{r+1} \left(e^{ir \pi /2} - 1\right)}{r^2} - i \dfrac{\pi^2}{16}\\ &= \dfrac{1}{i} \left( iG - \dfrac{\pi^2}{16} \right) - i \dfrac{\pi^2}{16} \\ &= \boxed{G} \end{align*}\]

where, \(G\) is Catalan's constant.

If you have an easy way, please do post it.

Substitute \(\displaystyle \tan x \mapsto x\)

\(\displaystyle \implies \text{I} = \int_{0}^{\infty} \dfrac{x^{7}\sin x}{(1+x^2)^4} \mathrm{d}x\)

From solution of Problem 28, we have,

\(\displaystyle \int_{0}^{\infty} \dfrac{x\sin mx}{(a^2+x^2)^3} \mathrm{d}x = \dfrac{\pi b e^{-a}}{16a^3} (1+am) \quad \quad \quad \quad \tag{*} \)

Differentiating \((*)\) \(6\) times w.r.t. \(m\) , \(1\) time w.r.t. \(a\), at \(a=m=1\) and simplifying, we get,

\(\displaystyle \text{I} = \boxed{\dfrac{5 \pi}{96e}}\)

Lemma \[\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 } \]

Giving the substitution \(x^2=t\), the integral becomes:

\[\int _{ 0 }^{ \infty }{ x^{ a }\cdot e^{ -x^{ 2 } }dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ t^{ \frac { a-1 }{ 2 } }\cdot e^{ -t }dt } =\frac { \Gamma \left( \frac { a+1 }{ 2 } \right) }{ 2 } \]

So the required integral is

\[=\frac { 1 }{ 4 } \left[ \Gamma \left( \frac { 1 }{ 2 } \right) \psi \left( \frac { 1 }{ 2 } \right) \right] =-\frac { 1 }{ 4 } \sqrt { \pi } (\gamma +\ln{(4)})\]

To show that \(\psi(1/2)=-\gamma-\ln(4)\), I'll use \(\psi(a)=H_{a-1}-\gamma\)

\[\psi(1/2)=H_{-1/2} - \gamma\]

From the definition of harmonic numbers:

\[H_{-1/2}=H_{1/2} - 2\]

\[H_{1/2}=\int _{ 0 }^{ 1 }{ \frac { 1-\sqrt { x } }{ 1-x } dx } =2\int _{ 0 }^{ 1 }{ t\left( \frac { 1-t }{ 1-t^{ 2 } } \right) dt } \]

I used the substitution \(t=\sqrt{x}\) above

\[\int _{ 0 }^{ 1 }{ t\left( \frac { 1-t }{ 1-t^{ 2 } } \right) dt } =\int _{ 0 }^{ 1 }{ \frac{t}{1+t}dt } =\int _{ 0 }^{ 1 }{ 1-\frac{1}{1+t} dt}=1-\ln(2)\]

Hence

\[H_{-1/2}=H_{1/2} - 2=2\left(1-\ln \left(2\right)\right)-2=-\ln(4)\]

\[\psi(1/2)=-\ln(4) - \gamma\]

Substitute \(e^{-x} \mapsto \sin x\)

\(\displaystyle \implies \text {I} = \int_{0}^{\frac {\pi}{2}} x \cot x \ \mathrm{d}x \)

Using Integration by parts,

\(\displaystyle \text {I} = - \int_{0}^{\frac {\pi}{2}} \ln (\sin x) \ \mathrm{d} x \)

\(\displaystyle= \dfrac{\pi}{2} \ln 2\)

Using Integration By Parts, we have,

\(\displaystyle \text{I} = 1 + \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right) \dfrac{\mathrm{d}x}{x} \)

\(\displaystyle = 1 + \text{J} \) (Let)

\(\displaystyle \text{J} = \int_{0}^{\infty} \int_{0}^{\infty} \cos x \cdot e^{-ax} - \dfrac{e^{-ax}}{x+1} \mathrm{d}x \ \mathrm{d}a \)

\(\displaystyle = \int_{0}^{\infty} \dfrac{a}{a^2+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-ax}}{x+1}\mathrm{d}x \ \mathrm{d}a \)

Substitute \(ax \mapsto x\)

\(\displaystyle \implies \text{J} = \int_{0}^{\infty} \dfrac{a}{a^2+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a \)

Since \(\displaystyle \int_{0}^{\infty} e^{-x}\mathrm{d}x = 1\)

\(\displaystyle \implies \text{J} = \int_{0}^{\infty} \int_{0}^{\infty} \dfrac{ae^{-x}}{a^2+1} - \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a \)

\(\displaystyle = \int_{0}^{\infty} e^{-x} \int_{0}^{\infty} \dfrac{a}{a^2+1} - \dfrac{1}{x+a}\mathrm{d}a \ \mathrm{d}x \)

\(\displaystyle = \int_{0}^{\infty} e^{-x} \left[\ln\left(\dfrac{\sqrt{a^2+1}}{a+x}\right)\right]_{a=0}^{a \to \infty} \mathrm{d}x \)

\(\displaystyle =\int_{0}^{\infty} e^{-x} \ln x \ \mathrm{d}x \)

\(\displaystyle = -\gamma\)

\(\displaystyle \implies \text{I} = \boxed{1 - \gamma}\)

This is a simple case of Euler substitution. Let \(y = \sqrt{x + \sqrt{x^2 + 2}} \), then \[y^2 = x + \sqrt{x^2 + 2 } \Rightarrow 2y \cdot \dfrac{dy}{dx} = 1 + \dfrac x{\sqrt{x^2+2}}= \dfrac {y^2}{\sqrt{x^2+2}} \Rightarrow dx = \dfrac{2\sqrt{x^2+2}}y dy. \]

The integral becomes \[ \int y \cdot \dfrac{2\sqrt{x^2+2}}y dy = 2\int \sqrt{x^2 + 2} \, dy = 2 \int (y^2 - x) \, dy \]

Because \(y^2 = x + \sqrt{x^2 + 2}\), then \[(y^2-x)^2 = x^2 + 2 \Rightarrow y^4 - 2xy^2 = 2 \Rightarrow x = \dfrac{y^4-2}{2y^2} = \dfrac12 y^2 - \dfrac1{y^2} . \]

Continue: \[ 2 \int \left (y^2 - \left(\dfrac12 y^2 - \dfrac1{y^2}\right) \right) \, dy = 2 \int \left (\dfrac12 y^2 + \dfrac1{y^2} \right) \, dy = \int \left (y^2 + \dfrac2{y^2} \right) \, dy = \dfrac13 y^3 - \dfrac2y + C . \]

Substituting back gives \[ \dfrac13 \left( \sqrt{x + \sqrt{x^2 + 2}}\right)^3 - \dfrac2{ \sqrt{x + \sqrt{x^2 + 2}}} + C. \]

Take \( \displaystyle f(n) = \int _{ 0 }^{ \frac { \pi }{ 4 } }{ n(\cos ^{ 2n }{ \theta } -\sin ^{ 2n }{ \theta } )\tan { (2\theta ) } d\theta } \)

Take \(cos(2\theta)=x \), to get :

\(\displaystyle f(n) = \frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n }-{ (1-x) }^{ n } }{ x } dx } \)

We have to find \(\displaystyle \lim _{ n\rightarrow \infty }{ f(n) } \)

We will first find the functional equation :

\( \displaystyle \frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n }-{ (1-x) }^{ n } }{ x } dx } =\frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ \frac { { (1+x) }^{ n-1 }-{ (1-x) }^{ n-1 } }{ x } dx } +\frac { n }{ { 2 }^{ n+1 } } \int _{ 0 }^{ 1 }{ { (1+x) }^{ n-1 }+{ (1-x) }^{ n-1 }dx } \)

\( \displaystyle f(n) = \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 } \)

Let the limit be \(L\)

\( \displaystyle \lim _{ n\rightarrow \infty }{ f(n) } =\lim _{ n\rightarrow \infty }{ \frac { nf(n-1) }{ 2(n-1) } +\frac { 1 }{ 2 } } \)

Hence \(\displaystyle L = \dfrac{1}{2}L+\dfrac{1}{2} \)

\(\Rightarrow \boxed{L=1} \)

Take the substitution \(t= \tan x \).

\[\begin{align*} \int_{0}^{\pi/2} \dfrac{ \csc x \sqrt{\tan x}}{\sin x + \cos x} dx &= 2\int_{0}^{\pi/2} \dfrac{\sqrt{\tan x}}{(1+\tan x ) (\sin 2x)} dx \\ &= 2\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(2t)} dt \\ &= \int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt \end{align*}\]

Take \(t = tan^2 (y)\).

\[\int_{0}^{\infty} \dfrac{\sqrt{t}}{(1+t)(t)} dt = \int_{0}^{\pi /2} \dfrac{\tan y}{\sec^2 (y) \tan^2 (y)} (2 \tan(y) \sec^2 (y) dy) = 2 \int_{0}^{\pi/2}dy = \pi. \]

\[I = \int_{0}^{1} \ln(\Gamma(x)) dx = \int_{0}^{1} \ln(\Gamma(1-t)) dt\]

I used the substitution \(x+t=1\). But from Euler's reflection formula for Gamma function i.e.

\[\Gamma(x) \Gamma(1-x)= \pi \csc(\pi x) \implies \ln(\Gamma(1-x)) + \ln(\Gamma(x)) = \ln(\pi) + \ln(\csc(\pi x))\]

\[\begin{align*} I &= \int_{0}^{1} \ln(\Gamma(1-x)) dx\\ I &= -\int_{0}^{1} \ln(\Gamma(x)) dx + \int_{0}^{1} \ln(\pi) dx+ \int_{0}^{1} \ln(\csc(\pi x))dx \\ I &= -I + \ln(\pi) - \int_{0}^{1} \ln(\sin(\pi x))dx \\ 2I &= \ln(\pi) - \dfrac{1}{\pi} \int_{0}^{\pi} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \int_{0}^{\pi / 2} \ln(\sin(x))dx \\ 2I &= \ln(\pi) - \dfrac{2}{\pi} \left(-\dfrac{\pi}{2} \ln(2) \right) \\ 2I &= \ln(\pi) + \ln(2) \\ I &= \dfrac{1}{2} \ln(2 \pi) \end{align*} \]

In the above solution I used the result that \(\int_{0}^{\pi/2} \ln(\sin(x))dx = -\dfrac{\pi}{2} \ln(2) \). This is proved here as follows:

\[\begin{align*} J &=\int_{0}^{\pi/2} \ln(\sin(x))dx = \int_{0}^{\pi/2} \ln(\sin(\pi/2 - x))dx = \int_{0}^{\pi/2} \ln(\cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x))dx + \int_{0}^{\pi/2} \ln(\cos(x))dx\\ 2J &= \int_{0}^{\pi/2} \ln(\sin(x) \cos(x))dx \\ 2J &= \int_{0}^{\pi/2} \ln(\sin(2x))dx - \int_{0}^{\pi/2} \ln(2)dx \\ 2J &= \dfrac{1}{2} \int_{0}^{\pi} \ln(\sin(x))dx - \dfrac{\pi}{2} \ln(2) \\ 2J &= \dfrac{1}{2}\left( 2\int_{0}^{\pi/2} \ln(\sin(x))dx \right) - \dfrac{\pi}{2} \ln(2) \\ 2J &= J - \dfrac{\pi}{2} \ln(2) \\ J &=- \dfrac{\pi}{2} \ln(2) \end{align*}\]

Define \(\displaystyle f(a)=\int _{ 0 }^{ \infty }{ \dfrac { ln(1+{ x }^{ a }) }{ (1+{ x }^{ 2 })ln(x) } dx } \)

Differentiating with respect to \(a\) we have :

\( \displaystyle f'(a)= \int _{ 0 }^{ \infty }{ \dfrac { { x }^{ a } }{ (1+{ x }^{ 2 })(1+{ x }^{ a }) } dx } \)

Use the substitution \( x=tan(\theta) \) we have :

\(\displaystyle f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \dfrac { \tan ^{ a }{ \theta } d\theta }{ (1+\tan ^{ a }{ \theta } ) } } \)

Using \( \displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f(x)dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ f(\dfrac { \pi }{ 2 } -x)dx } \) we have :

\( \displaystyle f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \dfrac { d\theta }{ (1+\tan ^{ a }{ \theta } ) } } \)

Using these two forms we have :

\( \displaystyle 2f'(a) = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ d\theta } =\dfrac { \pi }{ 2 } \)

\( \Rightarrow f'(a)=\dfrac { \pi }{ 4 } \)

Integrating this with upper limit as \(a\) and lower limit as \(b\) we have :

\( \displaystyle \int _{ 0 }^{ \infty }{ \dfrac { ln\left(\dfrac { 1+{ x }^{ a } }{ 1+{ x }^{ b } } \right) }{ (1+{ x }^{ 2 })ln(x) } dx } = \dfrac{\pi}{4} (a-b) \)

Put \( {x}^{2}=y \) to get the integral as :

\(\displaystyle I = \frac { 1 }{ 2 } \int { \frac { y{ e }^{ y } }{ { (y+1) }^{ 2 } } dy } \)

It can be also written as :

\( \displaystyle I = \frac { 1 }{ 2 } \int { \frac { { e }^{ y } }{ (y+1) } -\frac { { e }^{ y } }{ { (y+1) }^{ 2 } } dy } =\frac { 1 }{ 2 } \int { \frac { d(\frac { { e }^{ y } }{ y+1 } ) }{ dy } dy } \)

\( I = \dfrac { { e }^{ y } }{ 2(y+1) } +C=\dfrac { { e }^{ { x }^{ 2 } } }{ 2({ x }^{ 2 }+1) } +C \)