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

Question

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 $\alpha$?

Answer 1

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.

Answer 2

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.

Answer 3

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

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

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.

Answer 4

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

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

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

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