$$I=\int_0^{\pi/2}\frac{R^{\tan\theta}}{1 - R^2R^{2\tan\theta}}\,d\theta$$ This integral came up during a light transport calculation from a crystal of (constant) reflectivity $R$. I don't see an obvious solution for the definite or indefinite integral, but would be interested to see if there is one.
Integration of $\int_0^{\pi/2}\frac{R^{\tan\theta}}{1 - R^2R^{2\tan\theta}}\,d\theta$
2
$\begingroup$
integration
-
0Is the expression in the denominator as intended? (Normally, one would just write $R^{2 + 2 \tan \theta}$.) – 2017-01-31
-
0Wolfram can't solve it. Check your calculations if you sure about existence of solution. – 2017-01-31
-
0there is a small chance that this can be solved by the abel plana formula – 2017-01-31
-
0I think there is not enough symmetry to get a nice closed form, but the behaviour of such function for $R\gg 1$ or $R\ll 1$ is simple to compute through standard techniques. – 2017-01-31
1 Answers
0
By setting $\theta=\arctan t$ we get that our integral equals
$$ \frac{1}{R^2}\int_{0}^{+\infty}\frac{R^{-t}}{1+R^{-2t-2}}\cdot\frac{dt}{1+t^2}\tag{1}$$
and assuming $R\gg 1$ we have:
$$ \frac{R^{-t}}{1+R^{-2t-2}} = R^{-t}-R^{-3t-2}+R^{-5t-4}-R^{-7t-6}+\ldots \tag{2}$$
hence:
$$ \mathcal{L}\left(\frac{R^{-t}}{1+R^{-2t-2}}\right) = \frac{1}{s+\log R}-\frac{1}{R^2}\cdot\frac{1}{s+3\log R}+\frac{1}{R^4}\cdot\frac{1}{s+5\log R}-\ldots\tag{3}$$
and since $\mathcal{L}^{-1}\left(\frac{1}{1+t^2}\right)=\sin(s)$, by the properties of the Laplace transform $(1)$ equals:
$$\frac{1}{R^2}\int_{0}^{+\infty}\frac{\sin s}{s+\log R}\,ds + O\left(\frac{1}{R^4}\right) \tag{4}$$
hence by integration by parts $(1)$ equals $\color{red}{\large\frac{1}{R^2\log R}(1+o(1))}$.
In a similar way you may compute the asymptotics for $R\ll 1$.