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Does the generalised integral

$\int_{0}^{\pi}\frac{\sqrt x}{\sin x}dx$

converge or diverge?

The first thing I would do here is split it into two integrals

$$\int_0^\pi \frac{\sqrt x}{\sin{x}}dx=\int_0^{\frac{\pi}{2}} \frac{\sqrt x}{\sin{x}}dx+\int_{\frac{\pi}{2}}^\pi \frac{\sqrt x}{\sin{x}}dx$$

But then I am a bit stuck. I don't know if I now should compare it to something (and in that case what?), or if I should expand it with Taylor or something.

Mittens
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PythonDaniel
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    I'm pretty sure $\int_{\pi/2}^{\pi}\frac{\sqrt{x}}{\sin(x)}dx$ diverges. Notice $$\frac{\sqrt{x}}{\sin(x)}\geq \frac{\sqrt{\pi/2}}{\sin(x)}=\csc(x)\sqrt{\pi/2}>0$$ for $x\in[\pi/2,\pi)$. – Matthew H. Sep 15 '20 at 20:35

2 Answers2

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For $x$ close to $0$, $$\frac{\sqrt x}{\sin (x)}=\frac{1}{\sqrt{x}}+\frac{x^{3/2}}{6}+O\left(x^{7/2}\right)$$ so no problem.

But close to $x=\pi$ $$\frac{\sqrt x}{\sin (x)}=-\frac{\sqrt{\pi }}{x-\pi }-\frac{1}{2 \sqrt{\pi }}+\frac{\left(3-4 \pi ^2\right) }{24 \pi ^{3/2}} (x-\pi )+O\left((x-\pi )^2\right)$$ and, here, there is a major issue.

Amazingly, the problem could have been solved $\color{red}{1,400}$ years ago using $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.

This would give $$I(a)=\int_{0}^{a}\frac{\sqrt x}{\sin (x)}dx\simeq \int_{0}^{a} \frac{5 \pi ^2-4 (\pi -x) x}{16 (\pi -x) \sqrt{x}} dx$$ $$I(a)=\frac{5 \pi ^{3/2}}{8} \tanh ^{-1}\left(\frac{\sqrt{a}}{\sqrt{\pi }}\right)-\frac{a^{3/2}}{6}$$ and, as shown below, it is a decent approximation $$\left( \begin{array}{ccc} a & \text{approximation} & \text{exact} \\ 0.25 & 0.98828 & 1.00209 \\ 0.50 & 1.41108 & 1.42619 \\ 0.75 & 1.75095 & 1.76576 \\ 1.00 & 2.05704 & 2.07133 \\ 1.25 & 2.35188 & 2.36586 \\ 1.50 & 2.65145 & 2.66535 \\ 1.75 & 2.97141 & 2.98529 \\ 2.00 & 3.33164 & 3.34531 \\ 2.25 & 3.76309 & 3.77615 \\ 2.50 & 4.32461 & 4.33677 \\ 2.75 & 5.16168 & 5.17417 \\ 3.00 & 6.89988 & 6.92410 \end{array} \right)$$

Claude Leibovici
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  • That is actually very interesting! Amazing how math has been around for so long and how creative they were. I love the history of math, thanks a lot! – PythonDaniel Sep 17 '20 at 20:09
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The first integral in the split is convergent; the second intergral in the split howeverdiverges since

$$\int^{\pi}_{\pi/2}\frac{\sqrt{x}}{\sin x}\,dx\geq \sqrt{\pi/2}\int^\pi_{\pi/2}\frac{dx}{\sin x}$$ Since $$\frac{2}{\pi}\leq \frac{\cos x}{\pi/2-x}\leq 1,$$

$$ \begin{align} \int^{\pi}_{\pi/2}\frac{1}{\sin x}\,dx &=\int^{\pi/2}_0\frac{dx}{\sin(x+\pi/2)}=\int^{\pi/2}_0\frac{\pi/2 -x}{(\pi/2-x)\cos x}\,dx\\ &\geq \int^{\pi/2}_0\frac{dx}{\pi/2-x}=\int^{\pi/2}_0\frac{du}{u}=\infty \end{align} $$

Mittens
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