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 $\begin{aligned} \int_{\gamma_1} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x)^2}{(x+a)^2 + b^2}\,dx \\ \int_{\gamma_3} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x + 2\pi i)^2}{(x+a)^2 + b^2}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = \int_{\varepsilon}^R \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \end{aligned}$ 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 $\begin{aligned} \mathrm{Res}_{z = -a\pm ib} f(z) & = \frac{\Big(\ln\sqrt{a^2+b^2} + i\pi \mp i\tan^{-1}\frac{b}{a}\Big)^2}{\pm 2ib} \\ \mathrm{Res}_{z = -a+ib} f(z) + \mathrm{Res}_{z = -a-ib} f(z) & = \frac{-4i(\ln\sqrt{a^2+b^2}+i\pi)\tan^{-1}\frac{b}{a}}{2ib} \\ & = - \frac{2}{b}\tan^{-1}\frac{b}{a}(\ln\sqrt{a^2+b^2} + i\pi) \end{aligned}$ 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 $\begin{aligned} \ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big) & = 2\int_0^\infty \frac{t}{\big(t^2 + \frac14(x+1)^2\big)(e^{2\pi t} - 1)}\,dt \\ & = 8\int_0^\infty \frac{t}{(4t^2 + (x+1)^2)(e^{2\pi t} - 1)}\,dt \\ & = 2\int_0^\infty \frac{t}{(t^2 + (x+1)^2)(e^{\pi t}-1)}\,dt \end{aligned}$ and hence $\begin{aligned} \int_0^\infty \ln x & \left(\ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big)\right)\,dx \\ & = 2\int_0^\infty \left(\int_0^\infty \frac{\ln x}{(x+1)^2 + t^2}\,dx\right)\,\frac{t}{e^{\pi t}-1}\,dt \\ & = 2\int_0^\infty \left(\frac{1}{t} \tan^{-1}t \ln\sqrt{t^2 + 1}\right)\,\frac{t}{e^{\pi t}-1}\,dt \; = \; \int_0^\infty \frac{\tan^{-1}t \ln(t^2+1)}{e^{\pi t} - 1}\,dt \\ & = \tfrac12\ln^22 + \ln 2 \ln\pi - 1 \end{aligned}$ 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 $\begin{aligned} \zeta(s,\tfrac12) & = 2^{s-1}\left(1 + \tfrac{1}{s-1}\right) + 2\int_0^\infty \frac{(\frac14 + y^2)^{-\frac12s}\,\sin\big(s \tan^{-1}2y\big)}{e^{2\pi y}-1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^{s+1}\int_0^\infty \frac{(1 + 4y^2)^{-\frac12s} \, \sin\big(\tan^{-1}2y\big)}{e^{2\pi y} - 1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^s \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin\big(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \end{aligned}$ 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 $\begin{aligned} F''(s) & = \frac{\partial^2}{\partial s^2}\int_0^\infty \frac{(1 + t^2)^{-\frac12s} \, \sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \\ & = \int_0^\infty \frac{(1+t^2)^{-\frac12s}}{e^{\pi t} - 1}\left\{ \begin{array}{l} \tfrac14\big(\ln(1+t^2)\big)^2 \sin(s\tan^{-1}t) - \ln(1 + t^2) \tan^{-1}t \cos\big(s \tan^{-1}t\big) \\ - \big(\tan^{-1}t\big)^2 \sin\big(s \tan^{-1}t\big)\end{array}\right\}\,dt \end{aligned}$ 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:

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

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

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

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