# Integral of $\int_{y_1}^{y_2} \exp\left(\, -\alpha x\,\right)\, x \sqrt{1-x^2}{\rm d}x$

Does the following integral have a closed form solution? $$\int_{y_1}^{y_2} \exp\left(\, -\alpha x\,\right)\, x \sqrt{1-x^2}{\rm d}x$$ $$0< y_1 < 1$$ $$0< y_2 < 1$$

Or is there an approximation which works for large $\alpha$?

• It seems WA does not find any closed form... – Dmoreno Jul 31 '14 at 9:44
• One thing point out is that, if one takes $x=\sin \theta$, then the integral is $$\int_{0}^{\sin y}e^{-\alpha \cos \theta}\cos^2\theta \sin\theta\,d\theta$$ which can be obtained from $\int_0^{\sin y} e^{-\alpha \cos\theta}\,d\theta$ by partial derivatives wrt $\alpha$. Problem is that this integral is simpler in appearance, it doesn't seem to have a closed form either... – Semiclassical Jul 31 '14 at 13:21
• I have to ask where are all these integrals coming from? All very similar (granted very different solutions) but I must know :)? – Chinny84 Aug 4 '14 at 21:37
• :) This is related to my research which is taking me a lot of time. – Hesam Aug 4 '14 at 21:49
• You could also do that integration by means of the Laplace Transform. – Enthusiastic Engineer Aug 12 '14 at 20:02

This could help here. $$\int^{y_2}_{y_1}\mathrm{e}^{-\alpha x}x\sqrt{1-x^2}dx = -\frac{d}{d\alpha}\int\mathrm{e}^{-\alpha x}\sqrt{1-x^2}dx$$ using $x = \cos u$

then

$$\int^{y_2}_{y_1}\mathrm{e}^{-\alpha x}x\sqrt{1-x^2}dx = -\frac{d}{d\alpha}\int_{\cos^{-1}y_1}^{\cos^{-1}y_2}\mathrm{e}^{-\alpha \cos u}\sin^2 u du$$ here the last bit was edited due to @semiclassical keen eye :). now using $$-\sin^2 u = \cos^2 u - 1$$

we find $$\frac{d}{d\alpha}\int_{\cos^{-1}y_1}^{\cos^{-1}y_2}\mathrm{e}^{-\alpha \cos u}\left[\cos^2 u -1\right]du = \frac{d}{d\alpha}\left[\frac{d^2}{d\alpha^2}-1\right]\int_{\cos^{-1}y_1}^{\cos^{-1}y_2}\mathrm{e}^{-\alpha \cos u}du$$

$\textbf{update}$ For the special case of $$\begin{eqnarray} \cos^{-1}y_1 &=& \pi/2,\\ \cos^{-1}y_2 &=& 0. \end{eqnarray}$$ which corresponds to choosing $(y_1,y_2) = (0,1)$ we obtain $$\int^{0}_{\pi/2}\mathrm{e}^{-\alpha \cos u}du = -\int_{0}^{\pi/2}\mathrm{e}^{-\alpha \cos u}du = -\frac{\pi}{2}\left[I_0(\alpha) - L_0(\alpha)\right]$$

thus $$\frac{d}{d\alpha}\left[\frac{d^2}{d\alpha^2}-1\right]\int_{\cos^{-1}y_1}^{\cos^{-1}y_2}\mathrm{e}^{-\alpha \cos u}du =\frac{d}{d\alpha}\left[\frac{d^2}{d\alpha^2}-1\right]\left(\frac{\pi}{2}\left[I_0(\alpha) - L_0(\alpha)\right]\right)$$

$$\begin{eqnarray} \frac{d}{d\alpha}\left[\frac{d^2}{d\alpha^2}-1\right]I_{0}(\alpha) &=& \frac{d^2}{d\alpha^2}I_{1}(\alpha) - I_{1}(\alpha)\\ &=& \frac{d}{d\alpha}\left[\frac{1}{\alpha}I_1(\alpha)+I_2(\alpha)\right] -I_1(\alpha)\\ &=& I_3(\alpha) +\frac{3}{\alpha}I_2(\alpha)- I_{1}(\alpha) \end{eqnarray}$$ and $$\begin{eqnarray} \frac{d}{d\alpha}\left[\frac{d^2}{d\alpha^2}-1\right]L_{0}(\alpha) &=& \left[\frac{d^2}{d\alpha^2} - 1\right]\left(\frac{1}{2}\left[L_{-1}(\alpha) +L_1(\alpha) + \frac{1}{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}\right]\right)\\ &=& \left[\frac{d^2}{d\alpha^2}-1\right]g(\alpha) \end{eqnarray}$$ $$g'(\alpha) = \frac{1}{4}\left[L_{-2}(\alpha)+2L_0(\alpha) +L_2(\alpha) + \frac{2\alpha^{-1}}{\sqrt{\pi}\Gamma\left(\frac{1}{2}\right)}+\frac{\alpha}{2\sqrt{\pi}\Gamma\left(\frac{5}{2}\right)}\right],\\ g''(\alpha) = \frac{1}{8}\left[L_{-3}(\alpha) + 3L_{-1}(\alpha) + 3L_{1}(\alpha)+L_{3}(\alpha) + \frac{4\alpha^{-2}}{\sqrt{\pi}\Gamma\left(\frac{-1}{2}\right)} +\frac{2}{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}+\frac{2^{-2}\alpha^2}{\sqrt{\pi}\Gamma\left(\frac{7}{2}\right)}\right]$$

Thus for this special case $$\int^{1}_{0}\mathrm{e}^{-\alpha x}x\sqrt{1-x^2}dx =\\ -\frac{\pi}{2}\left[I_3(\alpha) +\frac{3}{\alpha}I_2(\alpha)- I_{1}(\alpha)-\frac{1}{8}\left[L_{-3}(\alpha) - L_{-1}(\alpha) - L_{1}(\alpha)+L_{3}(\alpha) + \frac{4\alpha^{-2}}{\sqrt{\pi}\Gamma\left(\frac{-1}{2}\right)} -\frac{2}{\sqrt{\pi}\Gamma\left(\frac{3}{2}\right)}+\frac{2^{-2}\alpha^2}{\sqrt{\pi}\Gamma\left(\frac{7}{2}\right)}\right]\right]$$ and subbing in for $\Gamma$'s we find $$-\frac{\pi}{2}\left[I_3(\alpha) +\frac{3}{\alpha}I_2(\alpha)- I_{1}(\alpha)-\frac{1}{8}\left[L_{-3}(\alpha) - L_{-1}(\alpha) - L_{1}(\alpha)+L_{3}(\alpha) -\frac{2\alpha^{-2}}{\pi} -\frac{8}{3\pi}+\frac{2\alpha^2}{15\pi}\right]\right]$$

ps. keep asking questions, hopefully I have not made another silly mistake.

• Very nice, I think you are close to solution. – Hesam Aug 4 '14 at 23:38
• Shouldn't $\sqrt{1-x^2}\,dx=\sin u \, d(\cos u)=\sin^2 u \, du$? (I noticed because it seemed too simple based upon my similar comment above.) – Semiclassical Aug 5 '14 at 0:18
• @Semiclassical yes you are totally right :)! I will edit accordingly. – Chinny84 Aug 5 '14 at 4:26
• It's a bit frustrating that $\int e^{-\alpha \cos \theta}\,d\theta$ isn't a nicer object. – Semiclassical Aug 5 '14 at 10:48
• @Semiclassical you telling me!! It is relative nice just on its own, but taking the derivatives has bound to create some error in the manipulation (3pages with mistakes included ha). But it troubles me that integrating over a different interval doesn't exist.. – Chinny84 Aug 5 '14 at 11:03

Does the following integral have a closed form solution?

In terms of elementary functions ? No, it does not. However, for $y=\pm1$ a closed form does exist, but in terms of the special functions Bessel I and Struve L.

• One lesson from Lucian's answer: definite integrals are often more well-studied i.e. have special function representations. Indefinite integrals, if they can't be explicitly integrated, are harder to deal with. @hesam – Semiclassical Jul 31 '14 at 14:11
• Do you know the expression when y=1? – Hesam Aug 1 '14 at 18:49
• $I(\pm1)=\dfrac\pi a\bigg[\dfrac{L_0(a)\mp I_2(a)}2-\dfrac{L_1(a)}a\bigg]$. – Lucian Aug 1 '14 at 18:56

Hint:

$\int_{y_1}^{y_2}e^{-\alpha x}x\sqrt{1-x^2}~dx$

$=\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}e^{-\alpha\sin x}\sin x\sqrt{1-\sin^2x}~d(\sin x)$

$=\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}e^{-\alpha\sin x}\sin x\cos^2x~dx$

$=\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}e^{-\alpha\sin x}\sin x(1-\sin^2x)~dx$

$=\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}e^{-\alpha\sin x}\sin x~dx-\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}e^{-\alpha\sin x}\sin^3x~dx$

$=\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{(2n)!}dx-\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{(2n+1)!}dx-\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+3}x}{(2n)!}dx+\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+4}x}{(2n+1)!}dx$

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.

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

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

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

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

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

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

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

$\therefore\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+1}x}{(2n)!}dx-\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+2}x}{(2n+1)!}dx-\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n}\sin^{2n+3}x}{(2n)!}dx+\int_{\sin^{-1}y_1}^{\sin^{-1}y_2}\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}\sin^{2n+4}x}{(2n+1)!}dx$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\alpha^{2n}\cos^{2k+1}x}{(2n)!k!(n-k)!(2k+1)}\right]_{\sin^{-1}y_1}^{\sin^{-1}y_2}-\left[\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}x}{2^{2n+1}n!(n+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}\sin^{2k+1}x\cos x}{2^{2n-2k+1}n!(n+1)!(2k+1)!}\right]_{\sin^{-1}y_1}^{\sin^{-1}y_2}-\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^{k+1}(n+1)!\alpha^{2n}\cos^{2k+1}x}{(2n)!k!(n-k+1)!(2k+1)}\right]_{\sin^{-1}y_1}^{\sin^{-1}y_2}+\left[\sum\limits_{n=0}^\infty\dfrac{(2n+3)\alpha^{2n+1}x}{4^{n+1}n!(n+2)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(2n+3)(k!)^2\alpha^{2n+1}\sin^{2k+1}x\cos x}{4^{n-k+1}n!(n+2)!(2k+1)!}\right]_{\sin^{-1}y_1}^{\sin^{-1}y_2}$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{k+1}n!\alpha^{2n}\left((1-y_2^2)^{k+\frac{1}{2}}-(1-y_1^2)^{k+\frac{1}{2}}\right)}{(2n)!k!(n-k)!(2k+1)}-\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\sin^{-1}y_2-\sin^{-1}y_1)}{2^{2n+1}n!(n+1)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(k!)^2\alpha^{2n+1}\left(y_2^{2k+1}\sqrt{1-y_2^2}-y_1^{2k+1}\sqrt{1-y_1^2}\right)}{2^{2n-2k+1}n!(n+1)!(2k+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^{k+1}(n+1)!\alpha^{2n}\left((1-y_2^2)^{k+\frac{1}{2}}-(1-y_1^2)^{k+\frac{1}{2}}\right)}{(2n)!k!(n-k+1)!(2k+1)}+\sum\limits_{n=0}^\infty\dfrac{(2n+3)\alpha^{2n+1}(\sin^{-1}y_2-\sin^{-1}y_1)}{4^{n+1}n!(n+2)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(2n+3)(k!)^2\alpha^{2n+1}\left(y_2^{2k+1}\sqrt{1-y_2^2}-y_1^{2k+1}\sqrt{1-y_1^2}\right)}{4^{n-k+1}n!(n+2)!(2k+1)!}$

It seems there is a closed form but I have not worked on it yet. However you can have this nice approximation for large $\alpha$

$$I = \int_{0}^{y} x\sqrt{1-x^2} e^{-\alpha x}\sim_{\alpha \sim \infty} \frac{1}{\alpha^2}- {\frac { \left( \alpha\,y+1 \right) {{\rm e}^{ -\alpha\,y}}}{{\alpha}^{2}}}.$$

You can use Laplace's method. See my answer where I laid out the basic idea behind it. Note that you can better approximations if you want. Here is a special case for $\alpha=100, y=1$

$$0.00009997,\\ 0.0001000 .$$

which they correspond to evaluating the integral and the approximation respectively.

• As your numerical check indicates, you have an extra factor of $1/2$ in your asymptotic. – Antonio Vargas Jul 31 '14 at 13:07
• @AntonioVargas: You are right. It was a typo. Thanks for the comment. – Mhenni Benghorbal Jul 31 '14 at 13:09
• Could you explain how you come up with this solution? I looked into your link but didn't get it. – Hesam Jul 31 '14 at 17:41
• @MhenniBenghorbal I think you posted an approximate for the following integral $\int_{0}^{y} \exp\left(\, -\alpha x\,\right)\, 2x {\rm d}x$ – Hesam Aug 4 '14 at 19:57
• @Hesam:It is for $\int_{0}^{y}x\sqrt{1-x^2} e^{-\alpha x}dx$. – Mhenni Benghorbal Aug 4 '14 at 23:46