How find this integral $I=\int_{0}^{1}\sqrt{1-W^2(x)}dx$

How find this nice integral

$$I=\int_{0}^{1}\sqrt{1-W^2(x)}dx$$

where, $W(x)$ is Lambert W function

My try: let $$\sqrt{1-W^2(x)}=u\Longrightarrow W(x)=\sqrt{1-u^2}$$

and since $x=W(x)e^{W(x)}$

so

$$x=W^{-1}(\sqrt{1-u^2})=\sqrt{1-u^2}e^{\sqrt{1-u^2}}?$$ so $$dx=-ue^{1-u^2}\dfrac{1+\sqrt{1-u^2}}{\sqrt{1-u^2}}du$$ so $$I=\int_{1}^{a}-u^2e^{1-u^2}\dfrac{1+\sqrt{1-u^2}}{\sqrt{1-u^2}}du$$

where $a$ such

$$\sqrt{1-a^2}e^{\sqrt{1-a^2}}=1(a>0)$$

then I can't,Thank you very much .

• What's the origin of the question? Are you sure there's a closed form solution in terms of some kind? – Will Nelson Nov 18 '13 at 4:46
• It would make more sense if the upper integration limit were e instead of $1$, since $W(e)=1$. But even so, a closed form solution is unlikely. – Lucian Nov 20 '13 at 14:38
• @Lucian: At this time all double summation terms will be cancelled and related to modified bessel function of the first kind. – doraemonpaul Nov 22 '13 at 5:20

$\int_0^1\sqrt{1-W^2(x)}~dx$

$=\int_0^\Omega\sqrt{1-x^2}~d(xe^x)$

$=\int_0^\Omega e^xx\sqrt{1-x^2}~dx+\int_0^\Omega e^x\sqrt{1-x^2}~dx$

$=\int_0^{\sin^{-1}\Omega}e^{\sin x}(\sin x)\sqrt{1-\sin^2x}~d(\sin x)+\int_0^{\sin^{-1}\Omega}e^{\sin x}\sqrt{1-\sin^2x}~d(\sin x)$

$=\int_0^{\sin^{-1}\Omega}e^{\sin x}\sin x\cos^2x~dx+\int_0^{\sin^{-1}\Omega}e^{\sin x}\cos^2x~dx$

$=\int_0^{\sin^{-1}\Omega}e^{\sin x}\sin x(1-\sin^2x)~dx+\int_0^{\sin^{-1}\Omega}e^{\sin x}(1-\sin^2x)~dx$

$=\int_0^{\sin^{-1}\Omega}e^{\sin x}~dx+\int_0^{\sin^{-1}\Omega}e^{\sin x}\sin x~dx-\int_0^{\sin^{-1}\Omega}e^{\sin x}\sin^2x~dx-\int_0^{\sin^{-1}\Omega}e^{\sin x}\sin^3x~dx$

$=\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}x}{(2n)!}dx+\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n+1)!}dx+\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}x}{(2n)!}dx+\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+2}x}{(2n)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+3}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+3}x}{(2n)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+4}x}{(2n+1)!}dx$

$=\int_0^{\sin^{-1}\Omega}\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}x}{(2n)!}\right)dx+\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{2(n+1)\sin^{2n+1}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=1}^\infty\dfrac{2n\sin^{2n+2}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{2(n+1)\sin^{2n+3}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+4}x}{(2n+1)!}dx$

For $n$ is any natural number,

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

This result can be done by successive integration by parts.

Similarly, for $n$ is any non-negative integer,

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

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

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

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

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

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

Similarly, $\int\sin^{2n+3}x~dx=\sum\limits_{k=0}^{n+1}\dfrac{(-1)^{k+1}(n+1)!\cos^{2k+1}x}{k!(n-k+1)!(2k+1)}+C$

$\therefore\int_0^{\sin^{-1}\Omega}\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}x}{(2n)!}\right)dx+\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{2(n+1)\sin^{2n+1}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=1}^\infty\dfrac{2n\sin^{2n+2}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{2(n+1)\sin^{2n+3}x}{(2n+1)!}dx-\int_0^{\sin^{-1}\Omega}\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+4}x}{(2n+1)!}dx$

$=\left[x+\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}x\cos x}{4^{n-k+1}(n!)^2(2k-1)!}\right]_0^{\sin^{-1}\Omega}+\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(-1)^{k+1}(n+1)!\cos^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^{\sin^{-1}\Omega}-\left[\sum\limits_{n=1}^\infty\dfrac{x}{4^n(n-1)!(n+1)!}-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\sin^{2k+1}x\cos x}{4^{n-k}(n-1)!(n+1)!(2k+1)!}\right]_0^{\sin^{-1}\Omega}-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{2(-1)^{k+1}(n+1)!(n+1)\cos^{2k+1}x}{(2n+1)!k!(n-k+1)!(2k+1)}\right]_0^{\sin^{-1}\Omega}-\left[\sum\limits_{n=0}^\infty\dfrac{(2n+3)x}{4^{n+1}n!(n+2)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(2n+3)(k!)^2\sin^{2k+1}x\cos x}{4^{n-k+1}n!(n+2)!(2k+1)!}\right]_0^{\sin^{-1}\Omega}$

$=\sum\limits_{n=0}^\infty\dfrac{\sin^{-1}\Omega}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\Omega^{2k-1}\sqrt{1-\Omega^2}}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2(-1)^k(n+1)!(1-\Omega^2)^{k+\frac{1}{2}}}{(2n+1)!k!(n-k)!(2k+1)}-\sum\limits_{n=1}^\infty\dfrac{\sin^{-1}\Omega}{4^n(n-1)!(n+1)!}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\Omega^{2k+1}\sqrt{1-\Omega^2}}{4^{n-k}(n-1)!(n+1)!(2k+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{2(-1)^k(n+1)!(n+1)(1-\Omega^2)^{k+\frac{1}{2}}}{(2n+1)!k!(n-k+1)!(2k+1)}-\sum\limits_{n=0}^\infty\dfrac{(2n+3)\sin^{-1}\Omega}{4^{n+1}n!(n+2)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(2n+3)(k!)^2\Omega^{2k+1}\sqrt{1-\Omega^2}}{4^{n-k+1}n!(n+2)!(2k+1)!}$

I would make the substitution $u=W(x)$, so $x=ue^u$ and $dx=(u+1)e^u du$, yielding $$I=\int_{0}^{\Omega}(u+1)\sqrt{1-u^2}\cdot e^udu;$$ here $\Omega=W(1)\approx0.573$. Now expand the square root in powers of $u$: $$I=\int_{0}^{\Omega}(u+1)\left(\sum_{k=0}^{\infty}c_ku^{2k}\right)e^u du=\sum_{k=0}^{\infty}c_k\int_{0}^{\Omega}(u^{2k}+u^{2k+1})e^u du =\sum_{k=0}^{\infty}c_k u^{2k+1}\left(\Gamma(2k+1,-u)-\Gamma(2k+2,-u)\right)\big\vert_{0}^{\Omega}=\sum_{k=0}^{\infty}c_k\left(\Omega^{2k+1}\Gamma(2k+1,-\Omega)-\Omega^{2k+2}\Gamma(2k+2,-\Omega)\right)=\sum_{k=1}^{\infty}d_k\Omega^{k},$$ where $$c_k=\frac{1}{k!}\frac{d^k}{dx^k}\sqrt{1-x}=(-1)^k\left(\frac{3}{2}-k\right)_{k}$$ and $$d_k=(-1)^{k-1} c_{\lfloor{(k-1)/2}\rfloor}\Gamma(k,-\Omega).$$ Putting these together, $$I=\sum_{k=1}^{\infty}(-1)^{k-1+\lfloor{(k-1)/2}\rfloor}\left(\frac{3}{2}-\left\lfloor{\frac{k-1}{2}}\right\rfloor\right)_{\lfloor{(k-1)/2}\rfloor}\Gamma(k,-\Omega)\Omega^{k}.$$