Forming a series representation for $\tan^a(u)$

Question

As part of my attempt at solving this integral, I became stuck with resolve the following definite integral: \begin{equation} \int_0^{\pi/4}u^{b}\tan^a\left(u\right)\,\mathrm{d}u \nonumber \end{equation} Where $a,b \in \mathbb{R}^+$. I have no idea how to address this integral in it's current form. My initial thought was to employ a series representation for $\tan^a(u)$ however I have no idea how this could be done. So my question, is it possible to form a series based representation for $\tan^a(u)$?? and if so, what methods can be employed?

Just hoping for a starting point :-)

Answer 1

Based on series expansion, starting from the usual expansion of $\frac{\tan(x)} x$, you have (using the generalized binomial expansion) $$\left(\frac{\tan (u)}{u}\right)^a=1+\frac{a }{3}u^2+\left(\frac{a^2}{18}+\frac{7 a}{90}\right) u^4+\frac{\left(35 a^3+147 a^2+124 a\right) }{5670}u^6+O\left(u^8\right)$$

Trying with $a=\pi$ and $b=e$, this would give $0.04879$ while numerical integration would give $0.04961$. Adding one extra term would lead to $0.04936$.

If you write $$u^b\,\tan^a(u)=u^{a+b} \left(1+\sum_{n=1}^\infty c_n\, u^{2n}\right)$$ the first coefficients write $$\left( \begin{array}{cc} n & c_n \\ 1 & \frac{a}{3} \\ 2 & \frac{a^2}{18}+\frac{7 a}{90} \\ 3 & \frac{a^3}{162}+\frac{7 a^2}{270}+\frac{62 a}{2835} \\ 4 & \frac{a^4}{1944}+\frac{7 a^3}{1620}+\frac{3509 a^2}{340200}+\frac{127 a}{18900} \\ 5 & \frac{a^5}{29160}+\frac{7 a^4}{14580}+\frac{2269 a^3}{1020600}+\frac{2011 a^2}{510300}+\frac{146 a}{66825} \\ 6 & \frac{a^6}{524880}+\frac{7 a^5}{174960}+\frac{5567 a^4}{18370800}+\frac{10397 a^3}{10206000}+\frac{5269351 a^2}{3536379000}+\frac{1414477 a}{1915538625} \\ 7 & \frac{a^7}{11022480}+\frac{7 a^6}{2624400}+\frac{1649 a^5}{55112400}+\frac{14891 a^4}{91854000}+\frac{4683409 a^3}{10609137000}+\frac{6470969 a^2}{11493231750}+\frac{32764 a}{127702575} \end{array} \right)$$

Answer 2

Substitution $$\tan u = x$$ gives: $$\int\limits_0^{\pi/4}u^b\tan^a u\,\mathrm du =\int\limits_0^1 x^a\arctan^b x \frac{\mathrm dx}{1+x^2} =\frac1{b+1}\int\limits_0^1 x^a\,\mathrm d\arctan^{b+1}x$$ $$=\frac1{b+1}x^a\arctan^{b+1}x\bigg|_0^1-\frac a{b+1}\int\limits_0^1 x^{a-1}\arctan^{b+1}x\,\mathrm dx,$$ $$=\frac1{b+1}\left(\frac\pi4\right)^{b+1}-\frac a{b+1}\int\limits_0^1 x^{a-1}\arctan^{b+1}x\,\mathrm dx,$$ wherein the Taylor series for the arctangent degrees at $x_0=1/\sqrt3$ has a good convergence.