Convergence of the improper integral $\int_{0}^{\pi/2}\tan^{p}(x) \; dx$

Question

I need help solving this integral: for which values of the parameter $p$ is this improper integral convergent?

$$\int_{0}^{\pi/2}\tan^{p}(x) \; dx$$

Thanks a lot!

Answer 1

Hint: substitute $u=\tan{x}$. Then $x=\arctan{u}$, $dx=\dfrac{du}{1+u^2}$, and the integral becomes,

$$\int_{0}^{\pi/2}\tan^{p}{x}\,dx=\int_0^{\infty}\dfrac{u^p}{1+u^2}du.$$

Answer 2

Hint

Continue using David H's suggestion. My contribution will be very minor : notice that
$$u^p=u^{p-2} [u^2 + 1 - 1]$$

Answer 3

The result seems to be $\frac{1}{2} \pi \sec \left(\frac{\pi p}{2}\right)$ provided that $-1<\Re(p)<1$

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{t \equiv {1 \over u^{2} + 1}\quad\iff\quad u = \pars{1 - t \over t}^{1/2}}$: \begin{align} \color{#00f}{\large\int_{0}^{\infty}{u^{p} \over u^{2} + 1}\,\dd u}&= \int_{1}^{0}t\,\pars{1 - t \over t}^{p/2}\,\half\,\pars{1 - t \over t}^{-1/2} \pars{-\,{\dd t \over t^{2}}} \\[3mm]&=\half\int^{1}_{0}t^{-\pars{p + 1}/2}\pars{1 - t}^{\pars{p - 1}/2}\,\dd t =\half\,{\rm B}\pars{-p + \half,p + \half} \\[3mm]&=\half\,{\Gamma\pars{-p + 1/2}\Gamma\pars{p + 1/2} \over \Gamma\pars{\bracks{-p + 1/2} + \bracks{p + 1/2}}} =\half\,{\pi \over \sin\pars{\pi\bracks{p + 1/2}}} = {\pi \over 2\cos\pars{\pi p}}\\[3mm]& = \color{#00f}{\large\half\,\pi\,\sec\pars{\pi p}}\,,\qquad \verts{\Re\pars{p}} < 1 \end{align}