Integrand tend to infinity at lower limit

Question

Can someone help me with integrating this: $$ \int_{0}^{\pi/2}\frac{dx}{\sqrt{\sin(x)}} $$ Here $$ \lim_{x\to0^+}\frac{1}{\sqrt{\sin(x)}}=\infty $$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}} & \,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\, \int_{0}^{1}t^{-1/2}\pars{1 - t^{2}}^{-1/2}\,\dd t \,\,\,\stackrel{t\ \mapsto\ t^{\large 1/2}}{=}\,\,\, {1 \over 2}\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t \\[5mm] & = {1 \over 2}\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}} = {1 \over 2}\,{\Gamma\pars{1/4}\root{\pi} \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}} \\[5mm] & = \bbx{{1 \over 4}\root{2 \over \pi}\,\Gamma^{2}\pars{1 \over 4}} \approx 2.6221 \end{align}

Answer 2

$$ \int_{0}^{\pi/2}\dfrac{1}{\sqrt{\sin x}}dx = \int_{0}^{\pi/2}\sin^{-1/2}x\cos^{0}xdx $$ But, $$ B(m,n) = 2\int_{0}^{\pi/2}\sin^{2m-1}x\cos^{2n-1}xdx $$ Thus, $$ \int_{0}^{\pi/2}\dfrac{1}{\sqrt{\sin x}}dx = \dfrac{1}{2}B(1/4,1/2) = \dfrac{\sqrt{\pi}\ \Gamma(1/4)}{\Gamma(3/4)} \simeq 2,6220576 $$

Answer 3

Notice that you can, utmost, use numerical methods to gain an approximate result.

For the sake of curiosity:

$$\frac{1}{\sqrt{\sin(x)}} \approx \frac{1}{\sqrt{x}} + \mathcal{O}(x^{3/2})$$

Truanting the series at the first order we then have

$$\int_0^{\pi/2} \frac{1}{\sqrt{\sin(x)}}\ \text{d}x \sim \int_0^{\pi/2} \frac{1}{\sqrt{x}}\ \text{d}x = 2\sqrt{x}\bigg|_0^{\pi/2} = \sqrt{2\pi} \approx 2.50663$$

A better result comes when you take more terms:

$$\frac{1}{\sqrt{\sin(x)}} \approx \frac{1}{\sqrt{x}} + \frac{x^{3/2}}{12} + \mathcal{O}(x^{7/2})$$

Whence

$$\int_0^{\pi/2} \frac{1}{\sqrt{\sin(x)}}\ \text{d}x \sim \int_0^{\pi/2} \frac{1}{\sqrt{x}}\ + \frac{x^{3/2}}{12}\text{d}x = \left(2\sqrt{x} + \frac{x^{5/2}}{30}\right)\bigg|_0^{\pi/2} = \sqrt{2 \pi }+\frac{\pi ^{5/2}}{120 \sqrt{2}} \approx 2.60971$$

More terms will give you the correct answer.

$$\frac{1}{\sqrt{\sin(x)}} \approx \frac{1}{\sqrt{x}}+\frac{x^{3/2}}{12}+\frac{x^{7/2}}{160}+\frac{61 x^{11/2}}{120960}+\frac{1261 x^{15/2}}{29030400}+O\left(x^{19/2}\right)$$

Answer 4

Using $$\sin(x)=\cos \left(\frac{\pi }{2}-x\right)=1-2\sin ^2\left(\frac{1}{2} \left(\frac{\pi }{2}-x\right)\right)$$ $$\int\frac{dx}{\sqrt{\sin (x)}}=\int\frac{dx}{\sqrt{1-2\sin ^2\left(\frac{1}{2} \left(\frac{\pi }{2}-x\right)\right)}}=-2\int \frac {dy}{\sqrt{1-2\sin^2(y)} }$$ leading to elliptic integrals and then $$\int\frac{dx}{\sqrt{\sin (x)}}=-2 F\left(\left.\frac{1}{2} \left(\frac{\pi }{2}-x\right)\right|2\right)$$ making $$\int_0^a\frac{dx}{\sqrt{\sin (x)}}=\sqrt{2} K\left(\frac{1}{2}\right)-2 F\left(\left.\frac{1}{4} (\pi -2 a)\right|2\right)$$ where appear the elliptic integral of the first kind $F(.)$ and the complete elliptic integral of the first kind $K(.)$ giving for $a=\frac \pi 2$ $$\int_0^{\frac \pi 2}\frac{dx}{\sqrt{\sin (x)}}=\frac{\pi ^{3/2}}{\sqrt{2} \,\Gamma \left(\frac{3}{4}\right)^2}$$ as already given in answers.

In the same spirit as Von Neumann's answer, we could also build quite good approximations using Padé approximants (built at $x=0$) instead of Taylor series. This could give for example $$\frac{1}{\sqrt{\sin (x)}}=\frac {1+\frac{1}{120}x^2 } {x^{1/2}-\frac{3 }{40}x^{5/2}}=\frac{1}{\sqrt{x}}-\frac{10 x^{3/2}}{9 \left( x^2-\frac{40}3\right)}$$ which is "almost" identical to a Taylor series to $O\left(x^{13/2}\right)$. Using $x=y^2$, this then leads to $$\int\frac{dx}{\sqrt{\sin (x)}}=-\frac 29 \int dy-\frac {800}{27}\int\frac {dy}{y^4-\frac{40}3}$$ Since $$\frac 1 {y^4-k^2}=\frac 1 {2k}\left(\frac 1{y^2-k}-\frac 1{y^2+k} \right)$$ $$\int\frac{dy} {y^4-k^2}=-\frac 1 {2k^{3/2}}\left(\tan ^{-1}\left(\frac{y}{\sqrt{k}}\right)+\tanh ^{-1}\left(\frac{y}{\sqrt{k}}\right)\right)$$ Using $k=\sqrt{\frac{40}{3}}$, this finally gives $$\int_0^a\frac{dx}{\sqrt{\sin (x)}}=\frac{2}{27} \left(5 \sqrt[4]{1080} \left(\tan ^{-1}\left(\frac{\sqrt[4]{\frac{3}{5}} }{2^{3/4}}\sqrt{a}\right)+\tanh ^{-1}\left(\frac{\sqrt[4]{\frac{3}{5}} }{2^{3/4}}\sqrt{a}\right)\right)-3 \sqrt{a}\right)$$ which, for $a=\frac \pi 2$ evaluates as $\approx 2.62189$ quite close to the exact solution.

Computed for a few values of $a$, here are some results $$\left( \begin{array}{ccc} a & \text{approximation} & \text{exact}\\ \frac{\pi }{20} & 0.79299 & 0.79299 \\ \frac{\pi }{10} & 1.12285 & 1.12285 \\ \frac{3 \pi }{20} & 1.37807 & 1.37807 \\ \frac{\pi }{5} & 1.59594 & 1.59594 \\ \frac{\pi }{4} & 1.79116 & 1.79116 \\ \frac{3 \pi }{10} & 1.97149 & 1.97149 \\ \frac{7 \pi }{20} & 2.14173 & 2.14174 \\ \frac{2 \pi }{5} & 2.30524 & 2.30527 \\ \frac{9 \pi }{20} & 2.46458 & 2.46465 \\ \frac{\pi }{2} & 2.62189 & 2.62206 \end{array} \right)$$

For sure, we could still improve using more terms $$\frac{1}{\sqrt{\sin (x)}}=\frac {1+\frac{1}{378}x^2-\frac{43 }{90720}x^4} { x^{1/2}-\frac{61 }{756}x^{5/2}}=\frac{1}{\sqrt{x}}+\frac{43 x^{3/2}}{7320}-\frac{35721 x^{3/2}}{610 \left(61 x^2-756\right)}$$ which is "almost" identical to a Taylor series to $O\left(x^{19/2}\right)$ and proceed the same way.

This last expansion would lead to $\approx 2.62203$ for the required integral.