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Background

I am studying the area of the implicit function curve

$$y^2=\sin \left(e^x\right).$$

I found that I need to calculate this integral:

$$I=\int_{-\infty }^{\ln \pi} \sqrt{\sin e^t} \, \rm{d}t.$$

Try 1

let $t = \sin e^x$, then

$$\int \frac{\sqrt{t}}{\sqrt{1-t^2} }\frac{1}{\arcsin t}\,\rm{d}t$$

I can't figure out how to calculate this.

Try 2

let $t = e^x$, then

$$\int_{0}^{\pi} \frac{\sqrt{\sin (t)}}{t}\,\rm{d}t$$

This form looks simple, but I still don't know how to calculate.

Question

Does $I$ have closed form?

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    $\begingroup$ I'm pretty sure that it doesn't have a closed form for the abstract integral. The numerical integral (on the specific interval) is a bit trickier $\endgroup$ May 13 at 2:56
  • $\begingroup$ @hellofriends. As an infinite series, no problem. $\endgroup$ May 13 at 4:32
  • $\begingroup$ It is also equal to: $$\pi\int_0^{\pi/2}\frac{\sqrt{\sin x}}{x(\pi-x)}\,\mathrm{d}x=\frac{\pi}{2}\int_0^\pi\frac{\sqrt{\sin x}}{x(\pi-x)}\,\mathrm{d}x$$If this proves useful I will add my derivation. I chose to push this representation because I think it’s more amenable to contour integration with more roots on the denominator but I’m not sure how to use this. $\endgroup$
    – FShrike
    May 13 at 8:06
  • $\begingroup$ Out of curiosity did you derive the implicit form from a set of parametric equations? $\endgroup$ May 13 at 23:32

1 Answer 1

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As you wrote $$I=\int_{-\infty }^{\log( \pi)} \sqrt{\sin (e^t)} \, dt=\int_{0 }^{\pi}\frac{\sqrt{\sin (x)}}{x}\,dx$$

Using series expansion $$\sqrt{\sin (x)}=\sum_{n=0}^\infty \frac {a_n}{b_n} x^{2 n+\frac{1}{2}} $$ The $a_n$ form the sequence $A008991$ and the $b_n$ form the sequence $A008992$ is $OEIS$.

So $$\int\frac{\sqrt{\sin (x)}}{x}\,dx=2\sum_{n=0}^\infty \frac {a_n}{(4n+1)\,b_n} x^{2 n+\frac{1}{2}}$$

The convergence is very slow

Edit

None of the CAS I was able to use gave the antiderivative or the integral. However, using the $\large 1,400$ years old approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician we can write $$\frac{\sqrt{\sin (x)}}{x} \simeq 2 \sqrt{\frac{(\pi -x)}{x (x-a) (x-b)}}\quad \text{with} \quad a=\left(\frac{1}{2}-i\right) \pi\quad \text{and} \quad b=\left(\frac{1}{2}+i\right) \pi$$ which can be integrated using elliptic integrals of the first and third kinds.

Just a few numbers for

$$I_k=\int_0^{k\frac{\pi}{12}}\frac{\sqrt{\sin (x)}}{x}\,dx$$

$$\left( \begin{array}{ccc} k & \text{approximation}& \text{solution} \\ 1 & 1.02920 & 1.02216 \\ 2 & 1.44820 & 1.44060 \\ 3 & 1.76172 & 1.75428 \\ 4 & 2.01665 & 2.00943 \\ 5 & 2.23049 & 2.22336 \\ 6 & 2.41171 & 2.40459 \\ 7 & 2.56500 & 2.55789 \\ 8 & 2.69317 & 2.68612 \\ 9 & 2.79790 & 2.79094 \\ 10 & 2.87997 & 2.87305 \\ 11 & 2.93881 & 2.93182 \\ 12 & 2.96882 & 2.96169 \end{array} \right)$$

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    $\begingroup$ @metamorphy. Thanks for pointing my stupidity. Cheers :-( $\endgroup$ May 13 at 5:14
  • $\begingroup$ Oh yes, $2\pi$ is a mistake $\endgroup$
    – GalAster
    May 13 at 5:41
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    $\begingroup$ Do the OEIS formulas have explicit forms? $\endgroup$ May 13 at 23:14

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