Brilliant Integration Contest - Season 3

Solution to problem 6 :

Let us denote the integral by \(I\) .

\(\displaystyle I = \int_0^\infty \frac{\ln x}{(x+a)^2+b^2}dx = \int_0^\infty \frac{\ln x}{(x+a+ib)(x+a-ib)}\)

With the substitution \(xy=a^2+b^2\) we have,

\(\displaystyle I=\ln(a^2+b^2)\int_0^\infty \frac{dy}{y^2+2ay+a^2+b^2} - \underbrace{\int_0^\infty \frac{\ln y}{(y+a)^2+b^2}dy}_I\)

\(\displaystyle I=\ln(\sqrt{a^2+b^2})\int_0^\infty \frac{dx}{(x+a)^2+b^2} = \frac{1}{b}\tan^{-1}\frac{b}{a}\ln\sqrt{a^2+b^2}\).

As an alternative to Aditya's proof...

Consider the cut plane \(\mathbb{C} \backslash [0,\infty)\), so that \(0 < \mathrm{Arg}\,z < 2\pi\) for all \(z\), and consider the integral of \[ f(z) \; = \; \frac{(\log z)^2}{(z+a)^2 + b^2} \] around the keyhole contour \(\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4\), where

• \(\gamma_1\) is the line segment \(z \,=\, x e^{0i}\), for \(\varepsilon < x < R\), just above the cut,
• \(\gamma_2\) is the circular contour \(z \,=\, Re^{i\theta}\) for \(0 < \theta < 2\pi\),
• \(\gamma_3\) the the line segment \(z \,=\, x e^{2\pi i}\), for \(\varepsilon < x < R\), just below the cut,
• \(\gamma_4\) is the circular contour \(z \,=\, \varepsilon e^{i\theta}\) for \(0 < \theta < 2\pi\).

where we assume that \(0 < \varepsilon < \sqrt{a^2+b^2} < R\). Now \[\begin{align} \int_{\gamma_1} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x)^2}{(x+a)^2 + b^2}\,dx \\ \int_{\gamma_3} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x + 2\pi i)^2}{(x+a)^2 + b^2}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = \int_{\varepsilon}^R \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \end{align}\] We also note that \[ \int_{\gamma_2} f(z)\,dz \; = \; O\big(\tfrac{\ln^2R}{R}\big) \hspace{2cm} \int_{\gamma_4} f(z)\,dz \; = \; O\big(\varepsilon \ln^2\varepsilon\big) \] as \(R \to \infty\) and \(\varepsilon \to 0\). Letting \(R \to \infty\) and \(\varepsilon \to 0\), we deduce that \[ \int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; 2\pi i\Big[ \mathrm{Res}_{z=-a+ib} f(z) \,+\, \mathrm{Res}_{z=-a-ib} f(z)\Big] \] Now \(f(z)\) has simple poles at each of \(-a \pm ib\), and \[\begin{align} \mathrm{Res}_{z = -a\pm ib} f(z) & = \frac{\Big(\ln\sqrt{a^2+b^2} + i\pi \mp i\tan^{-1}\frac{b}{a}\Big)^2}{\pm 2ib} \\ \mathrm{Res}_{z = -a+ib} f(z) + \mathrm{Res}_{z = -a-ib} f(z) & = \frac{-4i(\ln\sqrt{a^2+b^2}+i\pi)\tan^{-1}\frac{b}{a}}{2ib} \\ & = - \frac{2}{b}\tan^{-1}\frac{b}{a}(\ln\sqrt{a^2+b^2} + i\pi) \end{align}\] and hence \[ \int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; \frac{4\pi}{b}\tan^{-1}\frac{b}{a}\big(\pi - i\ln\sqrt{a^2+b^2}\big) \] Taking imaginary parts of this equation gives the result.

If we use the substitution, \(\displaystyle x=\frac{1}{u+1}\) then

\(\displaystyle I =\int_0^\infty \frac{u^{b-1}}{(up+p+1)^{a+b}}\,du\)

Again using the substitution, \(\displaystyle u=y\left(\frac{p+1}{p}\right)\) the integral transforms to,

\(\displaystyle I =\frac{1}{p^b(1+p)^a}\int_0^\infty \frac{y^{b-1}}{(1+y)^{a+b}}\,dy = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right)\)

This is a really nice use of previous results of this round of the contest!

Starting with the identity \[ \psi(x) \; = \; \ln x - \frac{1}{2x} - 2\int_0^\infty \frac{t}{(t^2+x^2)(e^{2\pi t}-1)}\,dt \] we deduce that \[\begin{align} \ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big) & = 2\int_0^\infty \frac{t}{\big(t^2 + \frac14(x+1)^2\big)(e^{2\pi t} - 1)}\,dt \\ & = 8\int_0^\infty \frac{t}{(4t^2 + (x+1)^2)(e^{2\pi t} - 1)}\,dt \\ & = 2\int_0^\infty \frac{t}{(t^2 + (x+1)^2)(e^{\pi t}-1)}\,dt \end{align}\] and hence \[\begin{align} \int_0^\infty \ln x & \left(\ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big)\right)\,dx \\ & = 2\int_0^\infty \left(\int_0^\infty \frac{\ln x}{(x+1)^2 + t^2}\,dx\right)\,\frac{t}{e^{\pi t}-1}\,dt \\ & = 2\int_0^\infty \left(\frac{1}{t} \tan^{-1}t \ln\sqrt{t^2 + 1}\right)\,\frac{t}{e^{\pi t}-1}\,dt \; = \; \int_0^\infty \frac{\tan^{-1}t \ln(t^2+1)}{e^{\pi t} - 1}\,dt \\ & = \tfrac12\ln^22 + \ln 2 \ln\pi - 1 \end{align}\] using the results of Problems \(5\) and \(6\).

Ishan, please set another one!

If we denote the integral by \(I\) & using \(2cosh(a\ln x)=x^a+x^{-a}\) & \(\displaystyle \ln(1+x)=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n}x^n\) the integral can be transformed into,

\(\displaystyle I = \sum_{n\ge 1} \frac{(-1)^{n-1}}{n} \int_0^1 (x^{n+a-1}+x^{n-a-1})=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n}(\frac{1}{n-a}+\frac{1}{n+a})=\frac{1}{2a}\sum_{n\ge 1} (-1)^{n-1} (\frac{1}{n-a}-\frac{1}{n+a})\)

For removing the alternating factor \((-1)^{n-1}\) this can be written as ,

\(\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a})-2\sum_{n\ge 1} (\frac{1}{2n-a}-\frac{1}{2n+a})\)

which simplifies to ,

\(\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a}-\frac{1}{n-\frac{a}{2}}+\frac{1}{n+\frac{a}{2}})\)

Using \(\displaystyle H_a = \sum_{n\ge 1}(\frac{1}{n}-\frac{1}{n+a})\) we have,

\(\displaystyle I=\frac{1}{2a}(H_a-H_{-a}-H_{\frac{a}{2}}+H_{-\frac{a}{2}})\)

Again using reflection formula for harmonic numbers as \(a<1\)

i.e. \(\displaystyle H_{1-a}-H_a = \pi \cot(\pi a)-\frac{1}{a}+\frac{1}{1-a}\) it simplifies to,

\(\displaystyle I = \frac{1}{2a} (-\pi\cot(\pi a)+\pi\cot(\pi\frac{a}{2})-\frac{1}{a}) = \frac{1}{2a}(\pi\csc(\pi a)-\frac{1}{a})\)

The integral can be written as : \(\displaystyle I=\underbrace{\int_0^1 x\tan^{-1} (3x) dx}_{I_1}+\underbrace{\int_0^3 \frac{\tan^{-1}x}{x}dx}_{I_2}\)

A simple integration by parts would yield \(\displaystyle I_1 = \frac{5}{9}\tan^{-1}3 - \frac{1}{6}\)

\(I_2\) is the Inverse Tangent Integral and we represent it by dilogarithms.

\(\displaystyle Ti_2(3)=\frac{1}{2i}(Li_2(3i)-Li_2(-3i))\)

So, \(\displaystyle I = \frac{5}{9}\tan^{-1}3 - \frac{1}{6} +\frac{1}{2i}(Li_2(3i)-Li_2(-3i))\)

Solution of Problem 5

Starting with the Hermite formula for the generalized Riemann zeta function: \[ \zeta(s,a) \; = \; \tfrac12a^{-s} + a^{1-s}(s-1)^{-1} + 2\int_0^\infty \frac{(a^2+y^2)^{-\frac12s} \, \sin\big(s \tan^{-1}\frac{y}{a}\big)}{e^{2\pi y} - 1}\,dy \] we have \[ \begin{align} \zeta(s,\tfrac12) & = 2^{s-1}\left(1 + \tfrac{1}{s-1}\right) + 2\int_0^\infty \frac{(\frac14 + y^2)^{-\frac12s}\,\sin\big(s \tan^{-1}2y\big)}{e^{2\pi y}-1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^{s+1}\int_0^\infty \frac{(1 + 4y^2)^{-\frac12s} \, \sin\big(\tan^{-1}2y\big)}{e^{2\pi y} - 1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^s \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin\big(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \end{align} \] so that \[ F(s) \; = \; \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \; = \; 2^{-s}\zeta(s,\tfrac12) - \frac{s}{2(s-1)} \] Now since \[ \begin{align} F''(s) & = \frac{\partial^2}{\partial s^2}\int_0^\infty \frac{(1 + t^2)^{-\frac12s} \, \sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \\ & = \int_0^\infty \frac{(1+t^2)^{-\frac12s}}{e^{\pi t} - 1}\left\{ \begin{array}{l} \tfrac14\big(\ln(1+t^2)\big)^2 \sin(s\tan^{-1}t) - \ln(1 + t^2) \tan^{-1}t \cos\big(s \tan^{-1}t\big) \\ - \big(\tan^{-1}t\big)^2 \sin\big(s \tan^{-1}t\big)\end{array}\right\}\,dt \end{align} \] we deduce that \[ F''(0) \; = \; -\int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \] and hence that the desired integral is \[ \int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -F''(0) \; = \; \frac{\partial^2}{\partial s^2}\Big[\tfrac{s}{2(s-1)} - 2^{-s}\zeta(s,\tfrac12)\Big] \Big|_{s=0} \] which simplifies to give \[ \int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -1 - \tfrac12\ln^22 + \ln2 \cdot \ln2\pi \; = \; -1 + \tfrac12\ln^22 + \ln2\ln\pi\]

Solution to Problem 8: \[\displaystyle I=\int_0^\frac{\pi}{2}\ln^2{a}+2\ln{a}\ln{\sin{x}}+\ln^2{\sin{x}}dx=\frac{\pi\ln^2{a}}{2}-\ln{a}\pi\ln{2}+\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx\]

\[\displaystyle\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx = \int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx\quad\text{(by }u= \sin{x})\]

\[\displaystyle\text{Now using }B(m,n) = 2\int_0^1x^{2m-1}(1-x^2)^{n-1}dx \text{ and thus }\int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx = \frac{d^2}{dm^2}\frac{B(m,\frac{1}{2})}{2*4}\big|_\frac{1}{2}\]

\[\displaystyle\text{Using }\Gamma'(m) = \Gamma(m)\psi(m),\qquad\Gamma''(m) = \Gamma(m)\psi(m)^2+\Gamma(m)\psi^{(1)}(m)\]

\[\displaystyle\frac{\sqrt{\pi}}{8}\frac{d^2}{dm^2}\frac{\Gamma(m)}{\Gamma(m+\frac{1}{2})}\Big|_\frac{1}{2} = \frac{\pi\ln^2{2}}{2}+\frac{\pi\zeta(2)}{4}\]

\[\displaystyle I=\boxed{\frac{\pi^3}{24}+\frac{\pi\ln^2{2}}{2}+\frac{\pi\ln^2{a}}{2}-\pi\ln{a}\ln{2}}\]

It's simplest using contour integration...

Note that \[\int_0^{2\pi} e^{\cos\theta}\,\cos(\sin\theta - n\theta)\,d\theta \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{\cos\theta} e^{i(\sin\theta - n\theta)}\,d\theta\right) \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{e^{i\theta} - in\theta}\,d\theta\right) \] With the substitution \(z = e^{i\theta}\), this becomes the contour integral \[ \mathfrak{Re}\left(\int_{|z|=1}e^z z^{-n} \frac{dz}{iz}\right) \; = \; \mathfrak{Re}\left(\frac{1}{i}\int_{|z|=1}\frac{e^z}{z^{n+1}}\,dz\right) \; = \; 2\pi\mathfrak{Re}\left(\mathrm{Res}_{z=0} \frac{e^z}{z^{n+1}}\right) \; = \; \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right. \]

\[ \text{I} = \int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{(\cos t - i \sin t)} \ \mathrm{d}t \right)} \]

\[ = \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{\Large e^{ -i t}} \ \mathrm{d}t \right)} \]

\[ = \dfrac{2 \pi}{n!} + \Re{ \left(\sum_{r \geq 0 \ ; \ r \neq n} \left( \dfrac{1}{r!} \int_{0}^{2 \pi} e^{i t (n-r)} \ \mathrm{d}t \right) \right)}\]

Since last integral is \(0\), we have,

\[ \text{I} = \dfrac{2 \pi}{n!} \]

If \( n \) is negative, we can write \(n = -k \ ; \ k \in \mathbb{Z}^+\), so that,

\[ \text{J} = \int_{0}^{2 \pi} e^{\cos t} \cos (kt + \sin t) \ \mathrm{d}t \]

By the above method, we can similarly show that \( \text{J} = 0 \).

Therefore,

\[ \int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right.\]

Solution to Problem 3:

\(\displaystyle H_n^{(k)} = \sum_{j=1}^{\infty}\frac{H_n^{(k)}x^i}{i^n}\quad H(x) = \frac{-\ln{(1-x)}}{1-x}\)

\[\displaystyle I = \int_{0}^{\infty}\frac{\ln^2{(1+x)}\ln{x}}{x}dx = -\int_{0}^{1}\frac{\ln{(1+x)}\ln^2{x}}{1+x}dx=\quad\text{by parts}\]

\[\displaystyle -\int_{1}^{2}\frac{\ln^2{(x-1)}\ln{x}}{x}dx=\quad\text{(by } u = 1+x)\]

\[\displaystyle -\int_{\frac{1}{2}}^{1}\frac{\ln^2{(\frac{1}{x}-1)}\ln{\frac{1}{x}}}{x}dx=\int_{\frac{1}{2}}^{1}\frac{\ln{x}\ln^2{(1-x)}}{x}-\frac{2\ln^2{x}\ln{(1-t)}}{x}+\frac{\ln^3{x}}{x}dx\quad\text{(by }u = 1/x)\]

\[\displaystyle\int\frac{\ln{x}\ln^2{(1-x)}}{x} = \frac{\ln^2{x}}{2}\ln^2{(1-x)}+\int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx\]

\[\displaystyle \int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx = -\sum_{n=1}^{\infty}H_n\frac{d^2}{dn^2}\int x^ndx = -\sum_{n=1}^{\infty}H_n\big(\frac{x^{n+1}\ln^2{x}}{n+1}-\frac{2x^{n+1}\ln{x}}{(1+n)^2}+\frac{2x^{n+1}}{(1+n)^3}\big)=\]

Now use \(H_n = H_{n+1} - \frac{1}{n+1}\) to obtain:

\[\displaystyle -\Big(H_1(x)\ln^2{x}-Li_2(x)\ln^2{x}-2H_2(x)\ln{x}+2Li_3(x)\ln{x}+2H_3(x)-2Li_4(x)\Big)\]

\[\displaystyle\textbf{Next }-2\int\frac{\ln^2{x}\ln{(1-x)}}{x}dx = 2\sum_{n=1}^{\infty}\frac{1}{n}\frac{d^2}{dn^2}\int x^{n-1} =\]

\[\displaystyle 2\sum_{n=1}^{\infty}\frac{x^n\ln^2{x}}{n^2}-\frac{2x^n\ln{x}}{n^3}+\frac{2x^n}{n^4} = 2\big(Li_2(x)\ln^2{x}-2Li_3(x)\ln{x}+2Li_4(x)\big)\]

\[\displaystyle I = \frac{\ln^4{x}}{4}\Big|^1_{\frac{1}{2}} + \frac{\ln^2{x}}{2}\ln^2{(1-x)}\Big|^1_{\frac{1}{2}}-2H_3(1)+2Li_4(1)+\]

\[\displaystyle +H_1(\frac{1}{2})\ln^2{2}-Li_2(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}-2Li_3(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-2Li_4(\frac{1}{2})+\]

\[\displaystyle4Li_4(1)-2Li_2(\frac{1}{2})\ln^2{2}-4Li_3(\frac{1}{2})\ln{2}-4Li_4(\frac{1}{2}) = \]

\[\displaystyle\frac{-3ln^4{2}}{4}-2H_3(1)+H_1(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-6Li_4(\frac{1}{2})+6Li_4(1)-3Li_2(\frac{1}{2})\ln^2{2}-6Li_3(\frac{1}{2})\ln{2}\]

Now using this for the evaluation of \(H_n(x)\)

\[\displaystyle -2H_3(1) = \frac{-\pi^4}{36},\ln^2{2}H_1(\frac{1}{2})=\frac{\ln^4{2}}{2}+\ln^2{2}Li_2{\frac{1}{2}},2\ln{2}H_2{\frac{1}{2}} = 2\ln{2}\zeta(3)-\zeta(2)\ln^2{2},2H_3(\frac{1}{2})= \frac{\pi^4}{360}+\frac{\ln^4{2}}{12}-\frac{\zeta(3)\ln{2}}{4}+2Li_4(\frac{1}{2})\]

Adding all this together and using formulas for \(Li_3\) and \(Li_2\)(I have checked on a calculator) gets the desired answer. Anyone may post the next problem.

Solution to Problem 4:

\(\displaystyle I=\int^1_0\frac{Li_3^2(-x)}{x^2}{\rm d}x\)

\[ I=-\frac{Li_3^2(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2(-x)Li_3(-x)}{x^2}{\rm d}x=-\frac{9}{16}\zeta^2(3)-\frac{2Li_2(-x)Li_3(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2^2(-x)}{x^2}{\rm d}x-\int^1_0\frac{2Li_3(-x)\ln(1+x)}{x^2}{\rm d}x\]

\[ I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{2Li_2^2(-x)}{x}\Bigg |^1_0-\int^1_0\frac{4Li_2(-x)\ln(1+x)}{x^2}{\rm d}x+\frac{2Li_3(-x)\ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{2Li_3(-x)}{x(1+x)}{\rm d}x-\int^1_0\frac{2Li_2(-x)\ln(1+x)}{x^2}{\rm d}x\]

\[ I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x+\frac{6Li_2(-x) \ \ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{6Li_2(-x)}{x(1+x)}{\rm d}x+\int^1_0\frac{6\ln^2(1+x)}{x^2}{\rm d}x\]

\[ I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9} \ {2}\zeta(3)-3\zeta(2)\ln{2}+\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x-\frac{6\ln^2(1+x)}{x}\Bigg |^1_0+\int^1_0\frac{12\ln(1+x)}{x(1+x)}{\rm d}x\]

\[ I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}+6\zeta(2)-12\ln^2{2}+\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x\]

Now,

\[ I_1=\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x=6Li_2(-x)\ln(1+x)\Bigg |^1_0+\int^1_0\frac{6\ln^2(1+x)}{x}{\rm d}x\]

\(I_1=\frac{3}{2}\zeta(3)-3\zeta(2)\ln{2}\)

\[ I_2=\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x=2Li_3(-x)\ln(1+x)\Bigg |^1_0-\int^1_0\underbrace{\frac{2Li_2(-x) \ln(1+x)}{x} {\rm d}x}_{\displaystyle\small{2Li_2(-x){\rm d}Li_2(-x)}}\]

\(I_2=\frac{1}{4}\zeta^2(2)-\frac{3}{2}\zeta(3)\ln{2}\)

Therefore,

\[\boxed{\displaystyle \int_0^1 \frac{{\rm Li}_3^2(-x)}{x^2}\, dx = -\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln 2+6\zeta(3)-6\zeta(2)\ln 2+6\zeta(2)-12\ln^2 2}\]

Solution to Problem 10:

\[\displaystyle\int_0^1xe^{-x}\ln{x}-e^{-x}\ln{x}dx = \int_0^1e^{-x}\ln{x}dx+\int_0^1e^{-x}dx-\int_0^1e^{-x}\ln{x}dx = \boxed{1-\frac{1}{e}}\text{ (by IBP)}\]

That's very nice of you to compile the past problems! Thanks :)

really nice :)

thank you

Results used : \(\displaystyle \sum_{n\ge 0}\frac{1}{n+a} =-\psi(a)\), \(\displaystyle \int \psi(ax) = \frac{1}{a}\ln(\Gamma(ax))\)

Let \(\displaystyle F(a)=\int_0^1 \frac{x^{a-1}}{(1+x)\ln x}dx\) then by differentiating w.r.t to a,

\(\displaystyle F'(a)=\int_0^1 \frac{x^{a-1}}{1+x}dx = \sum_{n\ge 0}(-1)^n \frac{1}{n+a}\)

Now , \(\displaystyle \sum_{n\ge 0}(-1)^n \frac{1}{n+a} = \frac{1}{2}(\sum_{n\ge 0} \frac{1}{n+\frac{a}{2}}-\sum_{n\ge 0} \frac{1}{n+\frac{a+1}{2}}) = \frac{1}{2}(\psi(\frac{a+1}{2})-\psi(\frac{a}{2}))\)

So , \(\displaystyle F'(\alpha)-F'(\beta) = \frac{1}{2}(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha}{2}))-\frac{1}{2}(\psi(\frac{\beta+1}{2})-\psi(\frac{\beta}{2})) \)

Integrating we have, \(\displaystyle F(\alpha)-F(\beta) = \left(\ln(\frac{\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{1+\beta}{2})\Gamma(\frac{\alpha}{2})})\right)+C\)

Now since, \(F(1)=0\implies C=0\)

So \(\displaystyle F(\alpha)-F(\beta) = \left(\ln\frac{\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{1+\beta}{2})\Gamma(\frac{\alpha}{2})}\right)\)

There is some delicate interchanging of integrals and infinite sums needed here, so I will set this out in full. Fix \(p,q > 0\). For \(0 < \varepsilon < X\) we have \[ \int_{\varepsilon}^X \frac{\cos^k px - \cos^k qx}{x}\,dx \; =\; \left(\int_{p\varepsilon}^{pX} - \int_{q\varepsilon}^{qX}\right) \frac{\cos^kx}{x}\,dx \; =\; \int_{qX}^{pX} \frac{\cos^kx}{x}\,dx - \int_{q\varepsilon}^{p\varepsilon}\frac{\cos^kx}{x}\,dx \] which implies that \[ \left|\int_{\varepsilon}^X \frac{\cos^k pq - \cos^k qx}{x}\,dx \right| \; \le \; \frac{2|p-q|}{\mathrm{min}(p,q)} \hspace{1cm} 0 < \varepsilon < X\] Thus, if we define the functions \[ f_k(X) \; = \; \int_0^X \frac{\cos^k px - \cos^k qx}{x}\,dx \hspace{2cm} k \in \mathbb{N}\,,\, X > 0 \] then we have shown that \[ \big|f_k(X)\big| \; \le \; \frac{2|p-q|}{\mathrm{min}(p,q)} \hspace{2cm} k \in \mathbb{N}\,,\, X > 0 \] and hence the functions \(f_k(X)\) are uniformly bounded in both \(k\) and \(X\). Moreover, the well-known Frullani integrals \[ \int_0^\infty \frac{\cos ax - \cos bx}{x}\,dx \; = \; \ln\big(\tfrac{b}{a}\big) \hspace{2cm} \int_0^\infty \frac{\sin ax - \sin bx}{x}\,dx \; = \; 0 \] which hold whenever \(a\) and \(b\) are nonzero with the same sign, imply that \[ \int_0^\infty \frac{e^{irpx} - e^{irqx}}{x}\,dx \; =\; \ln\big(\tfrac{q}{p}\big) \hspace{2cm} r \neq 0 \] and hence, since \[ f_k(X) \; = \; 2^{-k} \int_0^X \frac{(e^{ipx} + e^{ipx})^k - (e^{iqx} + e^{-iqx})^k}{x}\,dx \; = \; 2^{-k} \sum_{r=0}^k {k \choose r}\int_0^X \frac{e^{i(2r-k)px} - e^{i(2r-k)qx}}{x}\,dx \] we deduce that \[ \varphi_k \; = \; \lim_{X \to \infty} f_k(X) \; = \; \left\{ \begin{array}{ll} \displaystyle 2^{-k} \sum_{r=0}^k {k \choose r} \ln\big(\tfrac{q}{p}\big) & k \mbox{ odd} \\ \displaystyle 2^{-k} \sum_{0 \le r \le k \atop r \neq \frac12k} {k \choose r}\ln\big(\tfrac{q}{p}\big) & k \mbox{ even} \end{array} \right. \] and hence that \[ \varphi_k \; = \; \left\{ \begin{array}{ll} \ln\big(\tfrac{q}{p}\big) & k \mbox{ odd} \\ \ln\big(\tfrac{q}{p}\big)\left(1 - 2^{-k}{k \choose \frac12k}\right) & k \mbox{ even} \end{array} \right. \] Now, for \(-1 < a < 1\), \[\begin{align} \int_0^X \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} & = \int_0^X \left\{ \ln\big(1 + \tfrac{2a}{1+a^2}\cos px\big) -\ln\big(1 + \tfrac{2a}{1+a^2}\cos qx\big)\right\}\,\frac{dx}{x} \\ & = \int_0^X \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k\big[\cos^k px - \cos^k qx\big]\,\frac{dx}{x} \\ & = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k f_k(X) \end{align} \] where the interchange of the sum and integral in the last line is valid because \(\big|\tfrac{2a}{a^2+1}\big| < 1\), and hence the series is uniformly convergent in \(x\). We can now use the uniform boundedness of the functions \(f_k(X)\) to deduce that \[ \begin{align} \int_0^\infty \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} & = \lim_{X \to \infty}\int_0^X \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} \\ & = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k \varphi_k \end{align} \] Thus the integral is equal to \[\begin{align} \ln\big(\tfrac{q}{p}\big)&\left[ \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k + \sum_{k=1}^\infty \frac{1}{2k}\big(\tfrac{2a}{a^2+1}\big)^{2k} 2^{-2k}{2k \choose k}\right] \\ & = \ln\big(\tfrac{q}{p}\big)\left[\ln\big(1 + \tfrac{2a}{a^2+1}\big) + \tfrac12\sum_{k=1}^\infty \frac{1}{k}{2k \choose k} \big(\tfrac{a^2}{(a^2+1)^2}\big)^{k}\right] \\ & = \ln\big(\tfrac{q}{p}\big)\left[\ln\big(\tfrac{(1+a)^2}{a^2+1}\big) - \ln\left(\tfrac12\left[1 + \sqrt{1 -4\tfrac{a^2}{(a^2+1)^2}}\right]\right)\right] \\ & = 2\ln\big(\tfrac{q}{p}\big)\ln(1+a) \end{align} \] as required.

Solution to Problem 19

\(\displaystyle I=\int_0^\infty (e^{-x^p}-e^{-x^q})d(\ln x)\)

\(\displaystyle I= [(e^{-x^p}-e^{-x^q})\ln x]_0^\infty + p\int_0^\infty e^{-x^p}x^{p-1}\ln x-q\int_0^\infty e^{-x^q}x^{q-1}\ln x\,dx\)

The substitutions \(x^p=u\;,x^q=v\) makes the last two integrals as,

\(\displaystyle p\int_0^\infty e^{-x^p}x^{p-1}\ln x-q\int_0^\infty e^{-x^q}x^{q-1}\ln x\,dx = \int_0^\infty e^{-x}x\ln x\, dx \left(\frac{q-p}{qp}\right)\)

\(\displaystyle = \int_0^\infty e^{-x}\ln x\, dx \left(\frac{q-p}{qp}\right) = \gamma\left(\frac{p-q}{qp}\right)\)

Now \(\displaystyle [(e^{-x^p}-e^{-x^q})\ln x]_0^\infty = \lim_{x\to\infty} \frac{\ln x}{\frac{1}{e^{-x^p}-e^{-x^q}}}- \lim_{x\to 0} \frac{\ln x}{\frac{1}{e^{-x^p}-e^{-x^q}}}\)

Both of these can be evaluated by applying L-Hospital's rule succesively and both of them are \(0\).

Hence \(\displaystyle \int_0^\infty \frac{(e^{-x^p}-e^{-x^q})}{x}\, dx = \gamma\left(\frac{p-q}{pq}\right)\)

Solution to Problem 19 :

Proposition 1 : \[ \int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1}\right)\dfrac{\mathrm{d}x}{x} = -\gamma \]

Proof : Let,

\[ \text{J} = \int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1}\right)\dfrac{\mathrm{d}x}{x} \]

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

\[\displaystyle = \int_{0}^{\infty} \dfrac{1}{a+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-ax}}{x+1}\mathrm{d}x \ \mathrm{d}a \]

Substitute \(ax \mapsto x\) in the second integral,

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

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

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

\[\displaystyle = \int_{0}^{\infty} e^{-x} \int_{0}^{\infty} \dfrac{1}{a+1} - \dfrac{1}{x+a}\mathrm{d}a \ \mathrm{d}x \]

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

\[\displaystyle =\int_{0}^{\infty} e^{-x} \ln x \ \mathrm{d}x \]

\[\displaystyle = -\gamma \quad \square\]

Proposition 2 : \[\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma \]

Proof : \[\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \int_{0}^{\infty} \int_{0}^{1} (1-e^{-x})e^{-yx} \mathrm{d}x \ \mathrm{d}y - \int_{0}^{\infty} \int_{1}^{\infty} e^{-x(y+1)} \mathrm{d}x \ \mathrm{d}y \]

\[\displaystyle = \int_{0}^{\infty} \left(\dfrac{1-e^{-y}}{y} + \dfrac{e^{-(y+1)} - 1}{y+1} \right) \mathrm{d}y - \int_{0}^{\infty} \dfrac{e^{(y+1)}}{y+1} \mathrm{d}y \]

\[\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y} \]

\[ = \gamma \ \left( \text{Using Proposition 1} \right) \ \square \]

Now,

\[\text{I} = \int_{0}^{\infty} \dfrac{e^{-x^p} - e^{-x^q}}{x} \ \mathrm{d}x \]

\[ = \left( \int_{0}^{1} \dfrac{1-e^{-x^q}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^q}}{x} \ \mathrm{d}x \right) - \left( \int_{0}^{1} \dfrac{1-e^{-x^p}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^p}}{x} \ \mathrm{d}x \right) \]

Substitute \( x^q \mapsto x \) in the first integral and \(x^p \mapsto x\) in the second integral to get,

\[ \text{I} = \dfrac{1}{q} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \ \mathrm{d}x \right) - \dfrac{1}{p} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \ \mathrm{d}x \right) \]

Using Proposition 2, we get,

\[ \text{I} = \left(\dfrac{1}{q} - \dfrac{1}{p}\right) \gamma \]

\[ \therefore \text{I} = \dfrac{p-q}{pq} \gamma \quad \square\]

Note that \[\begin{align} I \; = \; \int_0^1 \frac{x \,\ln x\, \tan^{-1}x}{1+x}\,dx & = \int_0^1 \frac{x \ln x}{1+x}\left(\int_0^x \frac{du}{1+u^2}\right)\,dx \\ & = \int_0^1 \left(\int_u^1 \frac{x \ln x}{1+x}\,dx \right)\,\frac{du}{1+u^2} \\ & = \int_0^1 \Big[x\ln x - x - \ln x \ln(1+x) - \mathrm{Li}_2(-x)\Big]_{x=u}^1\,\frac{du}{1+u^2} \\ & = \int_0^1 \Big(-1 + \tfrac{1}{12}\pi^2 - u\ln u + u + \ln u \ln(1+u) +\mathrm{Li}_2(-u)\Big]\,\frac{du}{1+u^2} \\ & = -\tfrac14\pi + \tfrac{1}{48}\pi^2 + \tfrac{1}{48}\pi^3 + \tfrac12\ln2 + \int_0^1 \frac{\ln u \ln(1+u) + \mathrm{Li}_2(-u)}{1 + u^2}\,du \end{align}\] It is possible (integrating by parts and so forth) to calculate the indefinite integrals of both parts of the remaining integral in terms of polylogarithms, but the end result is extremely complicated! My expression for the indefinite integral \[ \int \frac{\ln u \ln(1+u)}{1 + u^2}\,du \] has over \(30\) terms, and the indefinite integral for \[ \int \frac{\mathrm{Li}_2(-u)}{1+u^2}\,du \] is even more complicated. The definite integrals are \[\begin{align} \int_0^1 \frac{\ln u \ln(1+u)}{1 + u^2}\,du & = \tfrac{1}{128} \left(\begin{array}{l}11 \pi^3 - 256 G \ln2 + 10 i \pi^2 \ln2 + 12\pi (\ln2)^2 \\ - 8i(\ln2)^3 - 394i\mathrm{Li}_3(\tfrac12+\tfrac12i) + 210\zeta(3)\end{array}\right) \\ \int_0^1 \frac{\mathrm{Li}_2(-u)}{1+u^2}\,du & = \tfrac{1}{384}\left(\begin{array}{l}-35\pi^3 - 30 i \pi^2 \ln2 - 36 \pi (\ln2)^2 + 576G \ln2\\ + 24i(\ln2)^3 - 768i\mathrm{Li}_3(\tfrac12+\tfrac12i) + 630i\zeta(3)\end{array}\right) \end{align}\] so that \[ \int_0^1 \frac{\ln x \ln(1+x) + \mathrm{Li}_2(-x)}{1 + x^2}\,dx \; = \; -\tfrac{1}{192}\pi^3 - \tfrac12 G \ln2 \] and hence \[ I \; = \; -\tfrac14\pi + \tfrac{1}{48}\pi^2 + \tfrac{1}{64}\pi^3 + \tfrac12\ln2 - \tfrac12G \ln2 \] as required.

That was exhausting! Someone else can set the next one...

SOLUTION OF PROBLEM 20:

Alternative solution.

Let \[\displaystyle R(x)=\int_0^x \dfrac{t\ln t}{1+t}dt=\int_0^1 \dfrac{tx^2\ln(tx)}{1+xt}dt\]

Note that,

\[R(1)=\dfrac{\pi^2}{12}-1\]

(Taylor expansion of integrand)

\[\begin{align} \displaystyle I=\int_0^1 \dfrac{x\ln x \arctan x}{1+x}dx \displaystyle &=\Big[R(x)\arctan x\Big]_0^1 -\int_0^1 \dfrac{R(x)}{1+x^2}dx\\ \displaystyle &= \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1\int_0^1 \dfrac{tx^2\ln(tx)}{(1+tx)(1+x^2)}dtdx\\ \displaystyle &= \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1\int_0^1 \dfrac{tx^2\ln t}{(1+tx)(1+x^2)}dtdx-\int_0^1\int_0^1 \dfrac{tx^2\ln x}{(1+tx)(1+x^2)}dtdx\\ &\displaystyle =\dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1 \left[t\ln t\left(\dfrac{t\ln(1+x^2)}{2(1+t^2)}+\dfrac{\ln(1+tx)}{t(t^2+1)}-\dfrac{\arctan x}{1+t^2}\right)\right]_{x=0}^{x=1} dt-\\ &\displaystyle \int_0^1 \left[\dfrac{\ln x\Big(tx-\ln(1+tx)\Big)}{1+x^2}\right]_{t=0}^{t=1}dx\\ &=\displaystyle\dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\dfrac{\ln 2}{2}\int_0^1 \dfrac{t^2\ln t}{1+t^2}dt-\int_0^1 \dfrac{\ln t\ln(1+t)}{1+t^2}dt+\dfrac{\pi}{4}\int_0^1 \dfrac{t\ln t}{1+t^2}dt-\int_0^1 \dfrac{x\ln x}{1+x^2}dx+\\ &\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx\\ &=\displaystyle \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\dfrac{\ln 2}{2}\int_0^1 \dfrac{t^2\ln t}{1+t^2}dt+\left(\dfrac{\pi}{4}-1\right)\int_0^1 \dfrac{t\ln t}{1+t^2}dt\\ \end{align}\]

Using Taylor expansion of\[\dfrac{x^2\ln x}{1+x^2}\] and \[\dfrac{x\ln x}{1+x^2}\] one obtains,

\[\displaystyle \int_0^1 \dfrac{x^2\ln x}{1+x^2}dx=G-1\]

G being the Catalan constant.

\[\displaystyle \int_0^1 \dfrac{x\ln x}{1+x^2}dx=-\dfrac{\pi^2}{48}\]

Therefore,

\[\boxed{\displaystyle I=\dfrac{\pi^3}{64}+\dfrac{\pi^2}{48}-\dfrac{\pi}{4}-\dfrac{1}{2}G\ln 2+\dfrac{1}{2}\ln 2}\]

Solution to Problem 7:

Consider the integrals of \[ f(z) \; = \; \frac{1}{z^n + 1} \hspace{1cm} g(z) \; = \; \frac{\log z}{z^n + 1}\] around the contour \(\gamma_1 + \gamma_2 - \gamma_3\), where

• \(\gamma_1\) is the line segment \(z \,=\, x\), for \(0 < x < R\),
• \(\gamma_2\) is the circular contour \(z \,=\, Re^{i\theta}\) for \(0 < \theta < \frac{2\pi}{n}\),
• \(\gamma_3\) the the line segment \(z \,=\, x e^{\frac{2\pi i}{n}}\), for \(0 < x < R\).

where \(R > 1\). Now \[\begin{align} \int_{\gamma_1} f(z)\,dz & = \int_0^R \frac{1}{x^n+1}\,dx \\ \int_{\gamma_3} f(z)\,dz & = e^{\frac{2 \pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{1}{x^n + 1}\,dx \\ \int_{\gamma_1} g(z)\,dz & = \int_0^R \frac{\ln x}{x^n+1}\,dx \\ \int_{\gamma_3} g(z)\,dz & = e^{\frac{2 \pi i}{n}}\int_0^R \frac{\ln x + i\frac{\pi}{n}}{x^n + 1}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,g(z)\,dz & = -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{\ln x}{x^n + 1}\,dx - i\frac{\pi}{n}e^{\frac{2\pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx \end{align}\] Since \[ \int_{\gamma_2} f(z)\,dz \; = \; O\big(R^{1-n}\big) \hspace{2cm} \int_{\gamma_2} g(z)\,dz \; = \; O\big(\ln R \,R^{1-n}\big) \] as \(R \to \infty\), we deduce that (since both \(f(z)\) and \(g(z)\) just have simple poles at \(e^{\frac{\pi i}{n}}\) inside this contour) \[ \begin{align} \displaystyle -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^\infty \frac{1}{x^n + 1}\,dx & = \displaystyle 2\pi i \mathrm{Res}_{z = e^{\frac{\pi i}{n}}} f(z) \; = \; -\frac{2\pi i}{n}e^{\frac{\pi i}{n}} \\ \displaystyle -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{\ln x}{x^n + 1}\,dx - i\frac{\pi}{n}e^{\frac{2\pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx & =\displaystyle 2\pi i \mathrm{Res}_{z = e^{\frac{\pi i}{n}}} g(z) \; = \; \frac{2\pi ^2}{n^2}e^{\frac{\pi i}{n}} \end{align}\] and so \[ \int_0^\infty \frac{1}{x^n + 1}\,dx \; = \; \frac{\pi}{n}\mathrm{cosec}\,\frac{\pi}{n} \hspace{2cm} \int_0^\infty \frac{\ln x}{x^n + 1}\,dx \; = \; -\frac{\pi^2}{n^2}\, \cot\frac{\pi}{n}\,\mathrm{cosec}\frac{\pi}{n} \]

I will be using the Euler result that \[ \sum_{u=1}^\infty\frac{H_u}{u^2} \; = \; 2\zeta(3) \] Integrating by parts, \[ \begin{align} \int_0^1 \mathrm{Li}_2^2(x)\,dx & = \zeta(2)^2 - 2\int_0^1 \mathrm{Li}_1(x)\mathrm{Li}_2(x)\,dx \; = \; \zeta(2)^2 - 2\sum_{r,s=1}^\infty \frac{1}{rs^2}\int_0^1 x^{r+s}\,dx \\ & = \zeta(2)^2 - 2\sum_{r,s=1}^\infty \frac{1}{rs^2(r+s+1)} \; = \; \zeta(2)^2 - 2\sum_{r,s=1}^\infty\frac{1}{s^2(s+1)}\left(\frac{1}{r} - \frac{1}{r+s+1}\right) \\ &= \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_{s+1}}{s^2(s+1)} \; = \; \zeta(2)^2 - 2\sum_{s=1}^\infty H_{s+1}\left(\frac{1}{s^2} - \frac{1}{s(s+1)}\right) \\ &= \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_{s+1}}{s^2} + 2\sum_{s=1}^\infty \frac{H_{s+1}}{s(s+1)} \\ & = \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_s}{s^2} - 2\sum_{s=1}^\infty \frac{1}{s^2(s+1)} + 2\sum_{s=1}^\infty H_{s+1}\left(\frac{1}{s} - \frac{1}{s+1}\right) \\ & = \zeta(2)^2 - 4\zeta(3) - 2\sum_{s=1}^\infty \left(\frac{1}{s^2} - \frac{1}{s(s+1)}\right) + 2\lim_{S \to \infty}\left(\sum_{s=1}^S \frac{H_{s+1}}{s} - \sum_{s=2}^{S+1} \frac{H_s}{s}\right) \\ &= \zeta(2)^2 - 4\zeta(3) - 2\zeta(2) + 2 + 2\lim_{S \to \infty}\left(\tfrac32 + \sum_{s=2}^S \frac{1}{s(s+1)} - \frac{H_{S+1}}{S}\right) \\ &= \zeta(2)^2 - 4\zeta(3) - 2\zeta(2) + 6 \end{align} \] as required.

Consider the principal branch of the logarithm, so that the plane is cut along the negative real axis. Suppose that \(a > 0\). If we integrate \[ f(z) \; = \; \frac{(\log z)^2}{z^2 + a^2} \] around the contour \(\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4\), where

• \(\gamma_1\) is the straight line \(z = x\) for \(\varepsilon < x < R\),
• \(\gamma_2\) is the semicircular arc \(z = Re^{i\theta}\) for \(0 < \theta < \pi\),
• \(\gamma_3\) is the straight line \(z = xe^{i\pi}\) for \(\varepsilon < x < R\), just above the cut,
• \(\gamma_4\) is the semicircular arc \(z = \varepsilon e^{i\theta}\) for \(0 < \theta < \pi\),

where \(0 < \varepsilon < a < R\). Then \[ \int_{\gamma_1} f(z)\,dz \; = \; \int_\varepsilon^R \frac{\ln^2x}{x^2 + a^2}\,dx \hspace{2cm} \int_{\gamma_3} f(z)\,dz \; = \; -\int_\varepsilon^R \frac{(\ln x + i\pi)^2}{x^2 + a^2}\,dx \] while \[ \lim_{\varepsilon \to 0}\int_{\gamma_4} f(z)\,dz \; = \; \lim_{R \to \infty}\int_{\gamma_2} f(z)\,dz \; = \; 0 \] and hence we deduce that \[ \int_0^\infty \frac{\ln^2x + (\ln x + i\pi)^2}{x^2+a^2}\,dx \; = \; 2\pi i \mathrm{Res}_{z=ia} f(z) \; = \; \tfrac{\pi}{a}(\ln a + \tfrac12\pi i)^2 \] so that \[ \int_0^\infty \frac{\ln^2x}{x^2 + a^2}\,dx \; = \; \tfrac{\pi}{2a}\,\ln^2a + \tfrac{\pi^3}{8a} \hspace{2cm} \int_0^\infty \frac{\ln x}{x^2 + a^2}\,dx \; = \; \tfrac{\pi}{2a}\ln a \]

Now \[ \frac{\tan^{-1}x}{x} \; =\; \int_0^1 \frac{du}{1 + x^2u^2} \] and hence \[ \begin{align} \int_0^\infty \left(\frac{\tan^{-1}x \ln x}{x}\right)^2\,dx & = \int_0^1 \int_0^1 \left(\int_0^\infty \frac{\ln^2x}{(1 + x^2u^2)(1 + x^2v^2)}\,dx\right)\,du\,dv \\ & = \int_0^1 \int_0^1 \left(\int_0^\infty \ln^2x\left\{ \frac{1}{x^2 + u^{-2}} - \frac{1}{x^2 + v^{-2}}\right\}\,dx \right)\,\frac{du\,dv}{u^2-v^2} \\ & = \int_0^1 \int_0^1 \left\{ \tfrac12\pi u\,\ln^2u + \tfrac18\pi^3u - \tfrac12\pi v\, \ln^2v - \tfrac18\pi^3v\right\} \frac{du\,dv}{u^2 - v^2} \\ & = \tfrac12\pi \int_0^1 \int_0^1 \frac{u \ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv + \tfrac18\pi^3 \int_0^1 \int_0^1 \frac{du\,dv}{u+v} \\ & = \tfrac12\pi\left\{ -\tfrac16\pi^2 + 4\ln2 - \tfrac12\zeta(3)\right\} + \tfrac18\pi^3 \times 2\ln2 \\ & = \pi\left(\tfrac14\pi^2\ln2 - \tfrac{1}{12}\pi^2 + 2\ln2 - \tfrac14\zeta(3)\right) \end{align} \] I initially relied on Mathematica to evaluate the double integral \[ \int_0^1\int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv \] but here is a proper derivation.

\[ \begin{align} \int_0^1 \int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv & = 2\int_0^1 \,du \int_0^u\,dv \frac{u\ln^2u - v\ln^2v}{u^2-v^2} \\ & = 2\int_0^1\,du \int_0^u\,dv \frac{1}{u^2-v^2}\int_v^u (2\ln w + \ln^2w)\,dw \\ & = \int\int\int_{0 < v < w < u < 1}\frac{2\ln w + \ln^2w}{u}\left(\frac{1}{u+v} + \frac{1}{u-v}\right) \\ & = \iint_{0 < w < u < 1} \frac{2\ln w + \ln^2w}{u}\ln\left(\tfrac{u+w}{u-w}\right)\,dw\,du \end{align}\] Since it is easy to check (by differentiating back again) that \[ \int (2\ln w + \ln^2w)\ln\left(\tfrac{u+w}{u-w}\right)\,dw \; = \; \begin{array}{l} \displaystyle u \ln^2w \ln\big(1 + \tfrac{w}{u}\big) + w\ln^2w \ln\big(\tfrac{u+w}{u-w}\big) \\ \displaystyle + u\ln^2w\ln\big(1 - \tfrac{w}{u}\big) + 2u\ln w \mathrm{Li}_2\big(-\tfrac{w}{u}\big) + 2u \ln w \mathrm{Li}_2\big(\tfrac{w}{u}\big) \\ \displaystyle - 2u\mathrm{Li}_3\big(-\tfrac{w}{u}\big) - 2u\mathrm{Li}_3\big(\tfrac{w}{u}\big) + c\end{array} \] it follows that \[ \int_0^u (2 \ln w + \ln^2w)\ln\left(\tfrac{u+w}{u-w}\right)\,dw \; = \; \tfrac16 u \left(\pi^2 \ln u + 12 \ln2 \ln^2u - 3 \zeta(3)\right) \] and so \[ \int_0^1 \int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv \; = \; \tfrac16\int_0^1\left(\pi^2 \ln u + 12 \ln2 \ln^2u - 3 \zeta(3)\right)\,du \; = \; -\tfrac16\pi^2 + 4\ln2 - \tfrac12\zeta(3) \] as required.

Note that \[ \begin{align} B(z,w) & = \int_0^\infty \frac{t^{z-1}}{(1+t)^{z+w}}\,dt \; = \; \left(\int_0^1 + \int_1^\infty\right) \frac{t^{z-1}}{(1+t)^{z+w}}\,dt \\ & = \int_0^1 \frac{t^{z-1}}{(1+t)^{z+w}}\,dt + \int_1^0 \frac{t^{1-z}}{(1+t^{-1})^{z+w}} \big(-t^{-2}\,dt\big) \\ & = \int_0^1 \frac{t^{z-1} + t^{w-1}}{(1+t)^{z+w}}\,dt \end{align} \] for all \(\mathfrak{Re}\,z\,,\,\mathfrak{Re}\,w \,>\, 0\), so that (with the substitution \(t = e^{-2x}\)) \[ \begin{align} B(m+in,m-in) & = \int_0^1 \frac{t^{m+in-1} + t^{m-in-1}}{(1+t)^{2m}}\,dt \; = \; \int_0^1 \frac{t^{m-1}(t^{in} + t^{-in})}{(1 + t)^{2m}}\,dt \\ & = \int_\infty^0 \frac{e^{2(1-m)x} \big(e^{-2inx} + e^{2inx}\big)}{(1 + e^{-2x})^{2m}}\,(-2e^{-2x})\,dx \\ & = 4\int_0^\infty \frac{e^{-2mx} \cos2nx}{(1 + e^{-2x})^{2m}}\,dx \; = \; 4\int_0^\infty \frac{\cos 2nx}{(e^x + e^{-x})^{2m}}\,dx \\ & = 4^{1-m}\int_0^\infty \frac{\cos 2nx}{\cosh^{2m}x}\,dx \end{align} \] for any real \(n\) and \(m > 0\), so that \[ \begin{align} \int_0^\infty \frac{\cos 2nx}{\cosh^{2m}x}\,dx & = 4^{m-1} B(m+in,m-in) \; = \; \frac{4^{m-1}}{\Gamma(2m)} \Gamma(m+in) \Gamma(m-in) \\ & = \frac{\sqrt{\pi}}{2\Gamma(m)\Gamma(m+\frac12)} \big|\Gamma(m+in)\big|^2 \end{align} \] for any real \(n\) and any \(m > 0\), using the duplication formula \[ \Gamma(2m) \;= \; \tfrac{1}{\sqrt{2\pi}} 2^{2m - \frac12} \Gamma(m) \Gamma(m+\tfrac12) \]

Since,

\[ \sin x = \dfrac{e^{ix} - e^{-ix}}{2i} \]

We have,

\[ \sin^{2n+1} x = \dfrac{(-1)^n}{2^{2n}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r} \sin (2r+1)x \]

\[\implies \int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{(-1)^n}{2^{2n}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r} \int_{0}^{\infty} \dfrac{\sin (2r+1)x}{x} \ \mathrm{d}x \]

\[ = \dfrac{(-1)^n \pi}{2^{2n+1}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r} \]

\[ = \dfrac{(-1)^n \pi}{2^{2n+1}} \sum_{r=0}^{n} \left( (-1)^r \dbinom{2n}{r} - (-1)^{r-1} \dbinom{2n}{r-1} \right) \]

Clearly, the above sum telescopes. Evaluating it, we have,

\[ \int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{(-1)^n \pi}{2^{2n+1}} (-1)^n \dbinom{2n}{n} \]

\[ \therefore \int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{\pi}{2^{2n+1}} \dbinom{2n}{n} \ \forall \ n \in \mathbb{Z}^+ \]

\( \displaystyle \large \int_{0}^{\frac{\pi}{4}} ln(1+\tan(\frac{\pi}{4} - x)\, dx\)

\( \displaystyle \large \int_{0}^{\frac{\pi}{4}} ln(\frac{2}{1+\tan x}\,dx\)

if we denote the integral by \(I\). it is also \(\frac{\pi ln2}{4} -I\)

and so the integral is \(\frac{\pi ln2}{8} \).

Start with the substitution \(y = x^2 + 1\). Then \[\begin{align} I_{p}(x) & = \int \frac{x\,dx}{(1 + x^2)^{1-p}(2 + x^2)^{1 + p}} \; = \; \tfrac12\int \frac{dy}{y^{1-p}(1 + y)^{1+p}} \\ & = \tfrac12 \int \left(\frac{y}{1+y}\right)^{p-1}\,\frac{dy}{(1+y)^2} \end{align}\] Then the substitution \(z = \frac{y}{1+y}\) gives \[ I_p(x) \; = \; \tfrac12\int z^{p-1}\,dz \; = \; \tfrac{1}{2p}z^p + c \; = \; \tfrac{1}{2p}\left(\tfrac{x^2+1}{x^2+2}\right)^p + c \] In this case, we have \(p = \tfrac{1000}{2012} = \tfrac{250}{503}\), and hence \[ I_{\frac{250}{503}}(x) \; = \; \tfrac{503}{500} \left(\tfrac{x^2+1}{x^2+2}\right)^{\frac{250}{503}} + c \]

Start with \[ f(a,b) \; = \; \int_0^{\frac12\pi} \sin^{a-1}x \cos^{b-1}x\,dx \; = \; \tfrac12B\big(\tfrac12a,\tfrac12b\big) \; = \; \frac{\Gamma(\frac12a)\Gamma(\frac12b)}{2\Gamma(\frac12a+\frac12b)} \] for \(a,b > 0\). Then \[\begin{align} \frac{\partial f}{\partial a} & = \tfrac12 f(a,b)\Big[\psi\big(\tfrac12a\big) - \psi\big(\tfrac12a+\tfrac12b\big)\Big] \\ \frac{\partial^2 f}{\partial a^2} & = \tfrac14 f(a,b) \Big[ \psi'\big(\tfrac12a\big) - \psi'\big(\tfrac12a+\tfrac12b\big) + \big[\psi\big(\tfrac12a\big) - \psi\big(\tfrac12a+\tfrac12b\big)\big]^2\Big] \end{align}\] and so \[ g(b) \; = \; \frac{\partial^2 f}{\partial a^2}(1,b) \; = \; \tfrac18 A(b)B(b) \] where \[ A(b) \; = \; \frac{\Gamma(\frac12)\Gamma(\frac12b)}{\Gamma(\frac12+\frac12b)} \; = \; \frac{\sqrt{\pi}\Gamma(\frac12b)}{\Gamma(\frac12+\frac12b)} \hspace{1cm} B(b) \; = \; \psi'\big(\tfrac12\big) - \psi'\big(\tfrac12+\tfrac12b\big) + \big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]^2 \] Thus \[\begin{align} A'(b) & = \tfrac12A(b)\big[\psi\big(\tfrac12b\big) - \psi\big(\tfrac12+\tfrac12b\big)\big] \\ A''(b) & = \tfrac14A(b) \Big[ \psi'\big(\tfrac12b\big) - \psi'\big(\tfrac12 + \tfrac12b\big) + \big[\psi\big(\tfrac12b\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]^2\Big] \end{align}\] and hence \[ A(1) \; = \; \pi \hspace{1.5cm} A'(1) \; = \; -\pi \ln2 \hspace{1.5cm} A''(1) \; = \; \tfrac{1}{12}\pi^3 + \pi(\ln2)^2 \] Also \[\begin{align} B'(b) & = -\tfrac12\psi''\big(\tfrac12+\tfrac12b\big) - \big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]\psi'\big(\tfrac12+\tfrac12b\big) \\ B''(b) & = -\tfrac14\psi'''\big(\tfrac12+\tfrac12b\big) + \tfrac12\psi'\big(\tfrac12+\tfrac12b\big)^2 - \tfrac12\big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]\psi''\big(\tfrac12+\tfrac12b\big) \end{align}\] and hence \[ B(1) \; =\; \tfrac13\pi^2 + 4(\ln2)^2 \hspace{1.5cm} B'(1) \; = \; \tfrac13\pi^2\ln2 + \zeta(3) \hspace{1.5cm} B''(1) \; = \; -\tfrac{1}{360}\pi^4 - 2 \zeta(3)\ln2 \] and hence \[\begin{align} \int_0^{\frac12\pi} \ln^2(\sin x) \ln^2(\cos x)\,dx & = \frac{\partial^4 f}{\partial a^2 \partial b^2} \Big|_{a=b=1} \; = \; g''(1) \\ & = \tfrac18\big[A''(1)B(1) + 2A'(1)B'(1) + A(1)B''(1)\big] \\ & = \tfrac{1}{320}\pi^5 + \tfrac12\pi(\ln2)^4 - \tfrac12\pi\zeta(3)\ln2 \end{align}\]

Let,

\[\text{I} = \int_{0}^{\infty} e^{-ax^2 - \frac{b}{x^{2}}} \ \mathrm{d}x \]

\[ = \dfrac{1}{2} \int_{-\infty}^{\infty} e^{-ax^2 - \frac{b}{x^{2}}} \ \mathrm{d}x \]

Sunstituting \(x \mapsto \dfrac{x}{\sqrt{a}}\), we have,

\[ \text{I}= \dfrac{1}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{-x^2 - \frac{ab}{x^{2}}} \ \mathrm{d}x \]

\[ = \dfrac{e^{-2\sqrt{ab}}}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{- \left(x - \frac{\sqrt{ab}}{x} \right)^2 } \ \mathrm{d}x \]

Using Glasser's Master Theorem, we have,

\[ \text{I} = \dfrac{e^{-2\sqrt{ab}}}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{-x^2 } \ \mathrm{d}x \]

\[ = \dfrac{1}{2} \sqrt{\dfrac{\pi}{a}} e^{-2\sqrt{ab}} \quad \square\]