Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions

Question

Let

$$\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$$

How to show that it converges with no use of trigonometric functions?

(trivially, it is the anti-derivative of $\sin ^{-1}$ and therfore can be computed directly)

Answer 1

This is an improper integral and the only problem to treat is at $1$. We have

$$\frac{1}{\sqrt{1-s^2}}\sim_1\frac1{\sqrt2\sqrt{1-s}}$$ and since the integral $$\int_0^1\frac{ds}{\sqrt{1-s}}$$ is convergent then the given integral is also convergent.

Answer 2

Here is a proof in single-variable calculus language. On the interval $0\leq s \leq 1$, we have $$ \sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \leq \sqrt{2} \sqrt{1-s} $$ So $$ \frac{1}{\sqrt{1-s^2}} \geq \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-s}} $$ We would like to use the Comparison Test, comparing $\int_0^1 \frac{ds}{\sqrt{1-s}}$ to $\int_0^1 \frac{ds}{\sqrt{1-s}}=\int_0^1 \frac{du}{\sqrt{u}}$. The latter converges by the $p$-test ($p=1/2 < 1$).

Unfortunately, the comparison goes in the wrong direction to apply the Comparison Test. We need to instead bound $\sqrt{1-s^2}$ from below in order to bound $\frac{1}{\sqrt{1-s^2}}$ from above. So assume instead that $\frac{1}{2} \leq s \leq 1$. Then $$ \sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \geq \sqrt{\frac{3}{2}} \sqrt{1-s} $$ So $$ \frac{1}{\sqrt{1-s^2}} \leq \sqrt{\frac{2}{3}} \frac{1}{\sqrt{1-s}} $$ on this interval.

To put it together, you have $$ \begin{aligned} \int_0^1 \frac{ds}{\sqrt{1-s^2}} &= \int_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \int_{1/2}^1 \frac{ds}{\sqrt{1-s^2}}\\ &\leq \int_0^{1/2}\frac{ds}{\sqrt{1-s^2}} + \sqrt{\frac{2}{3}} \int_{1/2}^1 \frac{ds}{\sqrt{1-s}} \end{aligned} $$ The first integral is not improper and the second converges.

Answer 3

Here's an alternative approach:

Using the Euler substitution $t=\sqrt{\frac{1-s}{1+s}}\implies s=\frac{1-t^2}{1+t^2}$, the integral is transformed to

$$\begin{align} \int_{0}^{1}\frac{\mathrm{d}s}{\sqrt{1-s^2}} &=\int_{1}^{0}\frac{t+\frac{1}{t}}{2}\,\frac{(-4t)\,\mathrm{d}t}{(1+t^2)^2}\\ &=2\int_{0}^{1}\frac{\mathrm{d}t}{1+t^2},\\ \end{align}$$

which is not an improper integral.

Answer 4

By replacing $t$ with $1-t$ we get: $$\int_{0}^{1}\frac{dt}{\sqrt{1-t^2}}=\int_{0}^{1}\frac{dt}{\sqrt{t(2-t)}}\leq\int_{0}^{1}\frac{dt}{\sqrt{t}}=2.$$

Answer 5

here is a way to show that $\int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}$ without the use of trigonometric functions. i will use the fact the area of unit circle is $\pi.$ that is $\int_0^1 \sqrt{1-x^2}dx = {\pi \over 4}$ twice and integration by parts.

\begin{eqnarray} {\pi \over 4} & = & \int_0^1 \sqrt{1-x^2}\ dx \\ & = & \int_0^1{1-x^2 \over \sqrt{1-x^2}} \ dx \\ & = &\int_0^1 \frac{dx}{\sqrt{1-x^2}}- \int_0^1{x^2 \over \sqrt{1-x^2}} \ dx \\ & = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} + \int_0^1 x\ d \sqrt{1-x^2} \\ & = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} +{x \sqrt{1-x^2}} |_0^1 - \int_0^1\sqrt{1-x^2}\ dx \\ & = &\int_0^1 \frac{dx}{\sqrt{1-x^2}} - {\pi \over 4}. \end{eqnarray}

and that proves the claim $$ \int_0^1 \frac{dx}{\sqrt{1-x^2}} = {\pi \over 2}. $$

Answer 6

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large% \int_{0}^{1}{\dd s \over \root{1 - s^{2}}}}}^{\ds{\dsc{s}\ =\ \dsc{1 - t^{2}}}}\ =\ \int_{1}^{0}{-2t\,\dd t \over {\root{1 - \pars{1 - t^{2}}^{2}}}} =2\int_{0}^{1}{\dd t \over \root{2 - t^{2}}} \\[5mm]&< 2\int_{0}^{1}{\dd t \over \root{2 - 1^2}} = \color{#66f}{\large 2} \end{align}