The Integral of Multiple Tangent Functions

Question

I need help to find the numerical values to the precision at least $50$ digits (the closed forms if possible) for the following integrals

\begin{equation} {\large\mathscr{F}}\left(\alpha,\beta,\mu\right)=\int_0^{\Large\frac{\pi}{2}}\bigg[\tan x\arctan\big(\beta\tan (\mu\tan x)\big)-\tan x\arctan\big(\alpha\tan (\mu\tan x)\big)\bigg]\ dx\\ \end{equation}

and

\begin{equation} {\large\mathscr{I}}=\int_0^{\Large\frac{\pi}{2}}\cot\left(\frac{\cot x}{2}\right)\cot x\ dx\\ \end{equation}

Somehow, my Mathematica $9.0$ failed to find the numerical values. It showed up warning messages like these:

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >> 

or

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} =

I use this code to obtain the numerical value of an integral (perhaps you have a better code that you might want to share with me)

N[Integrate[(integrand), {x,a,b}], (digits precision)]

I am interested in knowing the numerical values of ${\large\mathscr{F}}\left(\alpha,\beta,\mu\right)$ for the specific values of the following variables: \begin{array}{|c|c|c|c|} \hline \text{No.} & \alpha & \beta & \mu \\[7pt] \hline 1 & 2 & 3 & 1 \\[7pt] \hline 2 & 3 & 5 & 2 \\[7pt] \hline 3 & \frac{3}{2} & 2 & \frac{1}{2}\\[7pt] \hline 4 & \frac{4}{3} & \frac{5}{3} & \frac{1}{3}\\[7pt] \hline 5 & \frac{5}{4} & \frac{3}{2} & \frac{1}{4}\\[7pt] \hline \end{array} Any help would be greatly appreciated. Thank you.

Edit :

As requested by Mr. Amzoti, I used this code (I use example values no. $1$ in the table)

N[Integrate[Tan[x] ArcTan[3 Tan[Tan[x]]] - Tan[x] ArcTan[2 Tan[Tan[x]]],{x,0,Pi/2}], 50]

Answer 1

For the second integral $$ \int_0^{\pi/2} \tan x\tan(2/\tan x)\,dx, $$ change the variables $$ \tan x = u, \qquad u = \frac1v $$ to get $$ \int_0^\infty \frac{dv}{v(1+v^2)}\tan 2v. $$ Then split the integration over the individual periods of $\tan$, getting the sum $$ \int_0^{\pi/4}ds\,\cot 2s \sum_{k\geq0}\left(\frac{1}{v_-(1+v_-^2)} - \frac{1}{v_+(1+v_+^2)} \right), \qquad v_\pm = \frac\pi2(k+\tfrac12)\pm s. $$ The sum can be done with a CAS in terms of polygamma functions, then the integral can be done numerically: $$ \begin{array}{rl} 1.5142909817&1206622699&9814154016&3412884691&1754296527&\\0554534308&1840600184&0711274192&8238561012&4123287312&\\7970541870&9009506626&9851358610&7548703861&4382369398&\ldots \end{array} $$

For the modified second integral using the same approach I get $$ 1.8283334379\ 3552390329\ 6350583941\ 3012496783\ 6378207215\ 1909448383\ldots $$

For the first integral, change the variable to $$ u = \tan x, \qquad u = \frac1\mu \left( \tfrac\pi2 + \pi s\right), \qquad s\geq-\tfrac12, $$ and use $$ \frac{u}{1+u^2} = \Re \frac{1}{i+u}, $$ so that the each individual part for $(\beta,\mu)$ can symbolically be written as the real part of $$ -\int_{-1/2}^\infty \frac\pi\mu\frac{\arctan(\beta\cot\pi s)}{i+u}\,ds. $$ Since $\cot\pi s$ has period $1$, split off the integral over $[-\tfrac12,0]$, and write the rest as a sum, regularized by the factor $z^k$, $|z|<1$, where the free parameter $z$ will be later taken as limiting $z\to 1$: $$ -\int_{-1/2}^0(\cdots) - \sum_{k\geq0}\int_0^1ds\,\arctan(\beta\cot\pi s)\frac{z^k \pi/\mu}{i+u+\pi k/\mu}. $$ Doing the sum over $k$, this is equal to $$ -\int_{-1/2}^0(\cdots) - \int_0^1 ds\,\arctan(\beta\cot\pi s)\Phi(z, 1, i\mu/\pi+\tfrac12+s), $$ where $\Phi$ is the Hurwitz Lerch phi function and using the limiting form $$ \Phi(z, 1, a) = -\gamma-\log(1-z)-\psi(a) + o(1), $$ ($\gamma$ is Euler's gamma, $\psi$ is digamma), the non-singular part of the integral is $$ -\int_{-1/2}^0(\cdots) - \int_0^1 ds\,\arctan(\beta\cot\pi s)(-\gamma-\Re\psi(i\mu/\pi+\tfrac12+s)) \qquad\qquad(*) $$ When two integrals $(\beta,\mu)$ and $(\alpha,\mu)$ are subtracted, the singular parts (in $z$, diverging as $\log(1-z)$ as $z\to1$) are identical: $$ \int_0^1\arctan(\beta\cot\pi s)\,ds = 0,$$ so only the non-singular parts need to be kept, which justifies this procedure. Even the $-\gamma$ term can be dropped.

The final integrals is easy to evaluate numerically as differences in $(*)$ for different values $(\beta,\mu)-(\alpha,\mu)$: $$ \begin{array}{rl} 0.0375498037&2982212925&4568723544&0398706302&0150535939&0201026802\ldots\\ 0.0048468107&7351794278&3854803548&8002752919&3912230233&5799187527\ldots\\ 0.0854502616&2875157796&1693832958&8743830124&3474549771&5684466950\ldots\\ 0.0961295058&0450574013&5986068340&3132551900&1795523024&9411717322\ldots\\ 0.0935379078&0619519591&4784481067&9022768695&3615022581&3494639786\ldots \end{array}$$

Answer 2

Hi Princess of Mathematics. Using Maple I am obtaining the following approximations for the first integral using the values of your table:

$$0.037363172275853$$ $$0.0050398726261960$$ $$0.086272634856537$$ $$0.097077534919905$$ $$0.093245471867952$$

All the best.

Answer 3

First, one may note that $$\int^\infty_0\frac{x\sin(ax)}{x^2+b^2}{\rm d}x=\operatorname*{Res}_{z=ib}\frac{\pi ze^{iaz}}{z^2+b^2}=\frac{\pi}{2}e^{-ab}$$ Applying the substitution $\mu\tan{x}\mapsto x$, we see that $\mathcal{I}$ is equal to \begin{align} \mathscr{F}(\alpha, \beta, \mu) =&\int^\beta_\alpha\int^\infty_0\frac{x}{x^2+\mu^2}\frac{\tan{x}}{1+\lambda^2\tan^2{x}}{\rm d}x\ {\rm d}\lambda\\ =&\int^\beta_\alpha\int^\infty_0\frac{x}{x^2+4\mu^2}\frac{\sin{x}}{1+\lambda^2+(1-\lambda^2)\cos{x}}{\rm d}x\ {\rm d}\lambda\\ \end{align} Observe that $$\sum^\infty_{n=1}a^n\sin(nx)=\Im\sum^\infty_{n=1}(ae^{ix})^n=\Im\frac{ae^{ix}}{1-ae^{ix}}=\frac{a\sin{x}}{1-2a\cos{x}+a^2}\ \ \ \text{(for $|a|<1$)}$$ Letting $a\mapsto\frac{b}{a}$, $$\frac{ab\sin{x}}{a^2-2ab\cos{x}+b^2}=\sum^\infty_{n=1}\left(\frac{b}{a}\right)^n\sin(nx)\ \ \ \text{(for $|b|<a$)}$$ Setting $a=\frac{\lambda+1}{\sqrt{2}}$ and $b=\frac{\lambda-1}{\sqrt{2}}$, then denoting $\xi=\frac{\lambda-1}{\lambda+1}$, we get $$\frac{\sin{x}}{1+\lambda^2+(1-\lambda^2)\cos{x}}=\frac{2}{\lambda^2-1}\sum^\infty_{n=1}\xi^n\sin(nx)$$ Integrating termwise, \begin{align} \mathscr{F}(\alpha, \beta, \mu) =&\int^\beta_\alpha\frac{2}{\lambda^2-1}\sum^\infty_{n=1}\xi^n\int^\infty_0\frac{x\sin(nx)}{x^2+4\mu^2}{\rm d}x\ {\rm d}\lambda\\ =&\int^\beta_\alpha\frac{\pi}{\lambda^2-1}\frac{e^{-2\mu}\xi}{1-e^{-2\mu}\xi}{\rm d}\lambda =\frac{\pi}{2}\int^\frac{\beta-1}{\beta+1}_\frac{\alpha-1}{\alpha+1}\frac{{\rm d}\xi}{e^{2\mu}-\xi}\\ \end{align} It follows that $$\Large{\mathscr{F(\alpha,\beta,\mu)}=\frac{\pi}{2}\ln\left(\frac{e^{2\mu}-\frac{\alpha-1}{\alpha+1}}{e^{2\mu}-\frac{\beta-1}{\beta+1}}\right)}$$