Integral of exponential with $x(1-x)$ term

Question

Does the following integral have a closed form solution?

$$ \int_{0}^{y} \exp\left(\,-\sqrt{\,x(1-x)\,}\,\right)\,{\rm d}x $$

Or must I settle with an approximation?

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Edit: Actual form of integral has an $\alpha$: $$ \int_{0}^{y} \exp\left(\, -\alpha \sqrt{\,x(1-x)\,}\,\right)\,{\rm d}x $$ I need the solution (or approximate solution) work with large $\alpha$ (and if just possible for small $\alpha$ too

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2nd Edit: Is there analytic expression for the definite integral containing the $\alpha$? $$ \int_{0}^{1} \exp\left(\, -\alpha \sqrt{\,x(1-x)\,}\,\right)\,{\rm d}x $$

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3rd Edit: Actually it is $-\alpha$ not $\alpha$

Answer 1

$\int_0^y e^{-\alpha\sqrt{x(1-x)}}~dx$

$=\int_0^y e^{-\alpha\sqrt{-(x^2-x)}}~dx$

$=\int_0^y e^{-\alpha\sqrt{-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)}}~dx$

$=\int_0^y e^{-\alpha\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}~dx$

$=\int_{-\frac{1}{2}}^{y-\frac{1}{2}}e^{-\alpha\sqrt{\frac{1}{4}-x^2}}~dx$

$=\int_\pi^{\cos^{-1}(2y-1)}e^{-\alpha\sqrt{\frac{1}{4}-\left(\frac{\cos x}{2}\right)^2}}~d\left(\dfrac{\cos x}{2}\right)$

$=\dfrac{1}{2}\int_{\cos^{-1}(2y-1)}^\pi e^{-\frac{\alpha\sin x}{2}}\sin x~dx$

$=\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{2^{2n+1}(2n)!}dx-\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{4^{n+1}(2n+1)!}dx$

For $n$ is any non-negative integer,

$\int\sin^{2n+2}x~dx=\dfrac{(2n+2)!x}{4^{n+1}((n+1)!)^2}-\sum\limits_{k=0}^n\dfrac{(2n+2)!(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}((n+1)!)^2(2k+1)!}+C$

This result can be done by successive integration by parts.

$\int\sin^{2n+1}x~dx$

$=-\int\sin^{2n}x~d(\cos x)$

$=-\int(1-\cos^2x)^n~d(\cos x)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}x~d(\cos x)$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\cos^{2k+1}x}{k!(n-k)!(2k+1)}+C$

$\therefore\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{2^{2n+1}(2n)!}dx-\int_{\cos^{-1}(2y-1)}^\pi\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{4^{n+1}(2n+1)!}dx$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\alpha^{2n}\cos^{2k+1}x}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}\right]_{\cos^{-1}(2y-1)}^\pi-\left[\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}x}{2^{4n+3}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}\sin^{2k+1}x\cos x}{2^{4n-2k+3}n!(n+1)!(2k+1)!}\right]_{\cos^{-1}(2y-1)}^\pi$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}+\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}(2y-1)(1-(2y-1)^2)^{k+\frac{1}{2}}}{2^{4n-2k+3}n!(n+1)!(2k+1)!}$

Answer 2

$\textbf{Possible direction}$

$$ \sqrt{x(1-x)} = \sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2} = \frac{1}{2}\sqrt{1-4\left(x-\frac{1}{2}\right)^2}. $$ thus making a change of variable $$ \cos(t) = 2\left(x-\frac{1}{2}\right),\\ -\sin(t)dt = 2dx. $$ we can re-write the integral as $$ -\frac{1}{2}\int_{\pi}^{0} \mathrm{e}^{\frac{\alpha}{2}\sin(t)}\sin(t)dt = \frac{1}{2}\int_{0}^{\pi} \mathrm{e}^{\frac{\alpha}{2}\sin(t)}\sin(t)dt $$

so the original integral looks like $$ \frac{1}{2}\int_{0}^{\pi}\mathrm{e}^{\frac{\alpha}{2}\sin(s)}\sin(s)ds = \frac{1}{2}\pi \left[L_{-1}\left(\frac{\alpha}{2}\right) + I_{1}\left(\frac{\alpha}{2}\right)\right]\approx 1.48983 $$ $\textbf{update:}$ There seems to be a discrepancy between this "answer" and the numerical result obtained in the comments above. If you could please refrain from voting until the differences are accounted for that will be great. Cheers!

$\textbf{update 2}$ The last equation has been modified from my previous answer due to my silly mistake of converting the limits of the integration! Anyway, using mathematica (apologies) the final integral is of the form modified Struve function $L_{-1}(x)$. The answer is approximately 1.48983 for $\alpha =1$ and limits 0 to 1, which corresponds to previous comment above.

$\textbf{update 3}$ Modified answer as mentioned in the comments to the original question. Now try to find a way to represent the special functions with negative arguments.

Answer 3

I'm sorry I'm not able to write good mathematical content on this site yet, I've just signed up.

I think it is possible to find an analytical solution to this. First the function : $x --> x*(1-x)$ is bijective when x is between 0 and $\frac{1}{2}$. So we can assume that x verifies that, or we seperate the integral in two, one part from 0 to $\frac{1}{2}$ and the other from $\frac{1}{2}$ to y knowing that the method to calculate it is the same.

First we change variables with $u^2 = x*(1-x)$ that gives $x= (1-\frac{\sqrt{1-4*u^2}}{2})$ (two solutions but only one of them is smaller than $\frac{1}{2}$))

replace it in the integral and use an integration by parts on $4*\frac{u}{\sqrt{(1-4*u^2)}}$ that should show up after the correct re-writing of the integral following the change of variable.

Then to simplify notation, use another one : $v=2*u$ ( we could do it before the integration by part)

If I'm right you should obtain a certain function of y (the main variable) which is made of roots and exponentials so is known) and $\int \sqrt{(1-v^2)}*e^{\frac{v}{2}}$ delimited by the proper values ( here 0 and $\frac{1}{2}*\sqrt{y(1-y)}$ ). Then you use a sin change of variable : v=sin(t) and it gives you :

the same term function of y, and $\int cos(t)^2*e^{\frac{sin(t)}{2}}$ From here, you use two integration by parts to sort the thing out and it should give you the analytical solution, ugly as it is :) . I may have been mistaken at some point so please don't hesitate to tell me if it turns out to be so. Thank you to those who read this to the end :)

Answer 4

For the range $0 \leq y \leq 1$, you may have a very accurate estimation of the integral expanding first the integrand as a Taylor series built at $x=0$. This gives $$e^{\sqrt{x(1-x)}}=1+\sqrt{x}+\frac{x}{2}-\frac{x^{3/2}}{3}-\frac{11 x^2}{24}-\frac{11 x^{5/2}}{30}-\frac{59 x^3}{720}-\frac{13 x^{7/2}}{630}+\frac{1513 x^4}{40320}-\frac{311 x^{9/2}}{22680}+\frac{14761 x^5}{3628800}-\frac{31417 x^{11/2}}{1247400}-\frac{594659 x^6}{479001600}-\frac{1877123 x^{13/2}}{97297200}-\frac{8409491 x^7}{87178291200}+O\left(x^{15/2}\right)$$ Then, for the integral, you have $$\int_{0}^{y} e^{\sqrt{x(1-x)}}\,{\rm d}x=y+\frac{2 y^{3/2}}{3}+\frac{y^2}{4}-\frac{2 y^{5/2}}{15}-\frac{11 y^3}{72}-\frac{11 y^{7/2}}{105}-\frac{59 y^4}{2880}-\frac{13 y^{9/2}}{2835}+\frac{1513 y^5}{201600}-\frac{311 y^{11/2}}{124740}+\frac{14761 y^6}{21772800}-\frac{31417 y^{13/2}}{8108100}-\frac{594659 y^7}{3353011200}-\frac{1877123 y^{15/2}}{729729000}-\frac{8409491 y^8}{697426329600}+O\left(y^{17/2}\right)$$ which is quite accurate.

Added later after OP's request

If we consider the case of $$\int_{0}^{y} e^{a\sqrt{x(1-x)}}\,{\rm d}x$$ assuming that $a$ is small, an identical procedure first leads to $$e^{a\sqrt{x(1-x)}}=1+a \sqrt{x}+\frac{a^2 x}{2}+\frac{1}{6} a \left(a^2-3\right) x^{3/2}+\frac{1}{24} a^2 \left(a^2-12\right) x^2+\frac{1}{120} a \left(a^4-30 a^2-15\right) x^{5/2}+\frac{1}{720} a^4 \left(a^2-60\right) x^3+\frac{a \left(a^6-105 a^4+315 a^2-315\right) x^{7/2}}{5040}+\frac{a^4 \left(a^4-168 a^2+1680\right) x^4}{40320}+\frac{a \left(a^8-252 a^6+5670 a^4+3780 a^2-14175\right) x^{9/2}}{362880}+\frac{a^6 \left(a^4-360 a^2+15120\right) x^5}{3628800}+\frac{a \left(a^{10}-495 a^8+34650 a^6-103950 a^4+155925 a^2-1091475\right) x^{11/2}}{39916800}+\frac{a^6 \left(a^6-660 a^4+71280 a^2-665280\right) x^6}{479001600}+\frac{a \left(a^{12}-858 a^{10}+135135 a^8-2702700 a^6-2027025 a^4+12162150 a^2-127702575\right) x^{13/2}}{6227020800}+\frac{a^8 \left(a^6-1092 a^4+240240 a^2-8648640\right) x^7}{87178291200}+O\left(x^{15/2}\right)$$ and, then, for the integral $$\int_{0}^{y} e^{a\sqrt{x(1-x)}}\,{\rm d}x=y+\frac{2}{3} a y^{3/2}+\frac{a^2 y^2}{4}+\frac{1}{15} a \left(a^2-3\right) y^{5/2}+\frac{1}{72} a^2 \left(a^2-12\right) y^3+\frac{1}{420} a \left(a^4-30 a^2-15\right) y^{7/2}+\frac{a^4 \left(a^2-60\right) y^4}{2880}+\frac{a \left(a^6-105 a^4+315 a^2-315\right) y^{9/2}}{22680}+\frac{a^4 \left(a^4-168 a^2+1680\right) y^5}{201600}+\frac{a \left(a^8-252 a^6+5670 a^4+3780 a^2-14175\right) y^{11/2}}{1995840}+\frac{a^6 \left(a^4-360 a^2+15120\right) y^6}{21772800}+\frac{a \left(a^{10}-495 a^8+34650 a^6-103950 a^4+155925 a^2-1091475\right) y^{13/2}}{259459200}+\frac{a^6 \left(a^6-660 a^4+71280 a^2-665280\right) y^7}{3353011200}+\frac{a \left(a^{12}-858 a^{10}+135135 a^8-2702700 a^6-2027025 a^4+12162150 a^2-127702575\right) y^{15/2}}{46702656000}+\frac{a^8 \left(a^6-1092 a^4+240240 a^2-8648640\right) y^8}{697426329600}+O\left(y^{17/2}\right)$$