# How can one show that $\int_{0}^{\pi/4}{\sqrt{\sin(2x)}\over \cos^2(x)}\mathrm dx=2-\sqrt{2\over \pi}\cdot\Gamma^2\left({3\over 4}\right)?$

Proposed:

$$\int_{0}^{\pi/4}{\sqrt{\sin(2x)}\over \cos^2(x)}\mathrm dx=2-\sqrt{2\over \pi}\cdot\Gamma^2\left({3\over 4}\right)\tag1$$

My try:

Change $(1)$ to

$$\int_{0}^{\pi/4}\sqrt{2\sec^2(x)\tan(x)}\mathrm dx\tag2$$

$$\int_{0}^{\pi/4}\sqrt{2\tan(x)+2\tan^3(x)}\mathrm dx\tag3$$

Not sure what substitution to use

How may we prove $(1)?$

• Perhaps not directly relevant, but $\int_0^{\frac\pi 2} \sqrt{\sin 2\theta} d\theta = \sqrt{\frac2\pi} \Gamma^2 (\frac 34)$ – B. Mehta May 18 '17 at 18:34
• Consider expanding $\sin(2x)=2\sin x\cos x$ to get an integral that's essentially $\sin^\frac12(x)\cos^{-\frac32}(x)$; Wolfram Alpha then suggests that this is basically an ${}_2F_1$ and your exponents mean that it should be a pretty straightforward special value. – Steven Stadnicki May 18 '17 at 18:34
• See also Wallis' integrals, especially the paragraph detailing their relationship to the beta and $\Gamma$ functions. – Lucian Aug 31 '17 at 0:18

By substituting $x=\arctan t$ our integral takes the form:
$$I=\int_{0}^{1}\sqrt{\frac{2t}{1+t^2}}\,dt$$ and by substituting $\frac{2t}{1+t^2}=u$ we get: $$I = \int_{0}^{1}\left(-1+\frac{1}{\sqrt{1-u^2}}\right)\frac{du}{u^{3/2}}$$ that is straightforward to compute through the substitution $u^2=s$ and Euler's Beta function: $$I = \frac{1}{2} \left(4+\frac{\sqrt{\pi }\,\Gamma\left(-\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right).$$ The identities $\Gamma(z+1)=z\,\Gamma(z)$ and $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ settle OP's $(1)$.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffe,10px]{\ds{\int_{0}^{\pi/4}{\root{\sin\pars{2x}} \over \cos^{2}\pars{x}}\,\dd x}} = \int_{0}^{\pi/4}{\root{\sin\pars{2x}} \over \bracks{1 + \cos\pars{2x}}/2}\,\dd x \\[5mm] \stackrel{2x\ \mapsto\ x}{=}\,\,\,& \int_{0}^{\pi/2}{\root{\sin\pars{x}} \over 1 + \cos\pars{x}}\,\dd x = \left.\Re\int_{x = 0}^{\pi/2}\pars{z - 1/z \over 2\ic}^{1/2} {1 \over 1 + \pars{z + 1/z}/2}\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.\root{2}\,\Im\int_{x = 0}^{\pi/2}\pars{{1 - z^{2} \over z}\,\ic}^{1/2} {\dd z \over \pars{1 + z}^{2}}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\ -\root{2}\,\Im\int_{1}^{\epsilon}\pars{1 + y^{2} \over y}^{1/2} \,{\ic\,\dd y \over \pars{1 + \ic y}^{2}} \\[3mm] &\ - \root{2}\,\Im\int_{\pi/2}^{0}{\exp\pars{\ic\bracks{\pi/4 - \theta/2}} \over \epsilon^{1/2}} \,\epsilon\expo{\ic\theta}\ic\dd\theta \\[3mm] &\ -\root{2}\,\Im\int_{\epsilon}^{1}\pars{1 - x^{2} \over x}^{1/2}\expo{\ic\pi/4}\,{\dd x \over \pars{1 + x}^{2}} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to} &\ \root{2}\,\int_{0}^{1}\pars{1 + y^{2} \over y}^{1/2} \,{1 - y^{2} \over \pars{1 + y^{2}}^{2}}\,\dd y - \int_{0}^{1}\pars{1 - x^{2} \over x}^{1/2}\,{\dd x \over \pars{1 + x}^{2}} \\[5mm] & = \underbrace{\root{2}\,\int_{0}^{1} \,{1 - y^{2} \over \pars{1 + y^{2}}^{3/2}}\,{\dd y \over y^{1/2}}} _{\ds{\mc{I}_{1}}}\ -\ \underbrace{\int_{0}^{1}{\pars{1 - x}^{1/2} \over \pars{1 + x}^{3/2}} \,{\dd x \over x^{1/2}}}_{\ds{\mc{I}_{2}}} = \,\mc{I}_{1} - \,\mc{I}_{2} \label{1}\tag{1} \end{align}
\begin{align} \mc{I}_{2} & \equiv \int_{0}^{1}{\pars{1 - x}^{1/2} \over \pars{1 + x}^{3/2}} \,{\dd x \over x^{1/2}} = \int_{0}^{1}x^{-1/2}\,\pars{1 - x}^{1/2}\, \bracks{1 - \pars{-1}x}^{\,-3/2}\,\dd x \\[5mm] & = \mrm{B}\pars{{1 \over 2},{3 \over 2}}\, {}_{2}\mrm{F}_{1}\pars{{3 \over 2},{1 \over 2};2;-1}\label{3}\tag{3} \end{align} $\ds{\mrm{B}}$ and $\ds{\,{}_{2}\mrm{F}_{1}\,}$ are the Beta and Hypergeometric Functions, respectively. \eqref{3} is the Euler Type Expression of the Hypergeometric Function. \begin{align} \mbox{Note that}\quad\mrm{B}\pars{{1 \over 2},{3 \over 2}} & = {\Gamma\pars{1/2}\Gamma\pars{3/2} \over \Gamma\pars{2}} = {\pi \over 2} \label{4}\tag{4} \end{align} The Hypergeometric function is evaluated with the Kummer Theorem. Namely, \begin{align} {}_{2}\mrm{F}_{1}\pars{{3 \over 2},{1 \over 2};2;-1} & = {\Gamma\pars{2}\Gamma\pars{7/4} \over \Gamma\pars{5/2}\Gamma\pars{5/4}} = {\pars{3/4}\Gamma\pars{3/4} \over \bracks{\pars{1/2}\pars{3/2}\root{\pi}}\bracks{\pars{1/4}\Gamma\pars{1/4}}} \\[5mm] & = {4\,\Gamma\pars{3/4} \over \root{\pi}}\, {1 \over \pi/\bracks{\Gamma\pars{3/4}\sin\pars{\pi/4}}} = {2 \over \pi}\,\root{2 \over \pi}\Gamma^{2}\pars{3 \over 4} \label{5}\tag{5} \end{align} With \eqref{4} and \eqref{5}, \eqref{3} becomes $$\bbx{\mc{I}_{2} \equiv \int_{0}^{1}{\pars{1 - x}^{1/2} \over \pars{1 + x}^{3/2}} \,{\dd x \over x^{1/2}} = \root{2 \over \pi}\Gamma^{2}\pars{3 \over 4}}$$ such that \eqref{1} becomes: $$\bbox[20px,#ffe,border:1px dotted navy]{\ds{% \int_{0}^{\pi/4}{\root{\sin\pars{2x}} \over \cos^{2}\pars{x}}\,\dd x = 2 - \root{2 \over \pi}\Gamma^{2}\pars{3 \over 4}}}$$