# Method of stationary phase, with Laplacian eigenvalue exponent

I'm trying to solve the following $$\int_{\mathbb{R}}dt\int_0^{\ell_p}dy_i\hat{\rho}(t)\mathrm{e}^{\left(\frac{it}{\hbar}\left(E-(\frac{\ell_p}{\pi})\sqrt{\sin^2(y_1\pi/\ell_p)+\sin^2(y_2\pi/\ell_p)+\sin^2(y_3\pi/\ell_p)+\sin^2(y_4\pi/\ell_p)}\right)\right)}$$ where $i=1,2,3,4$ and $E$, $\ell_p$ and $A$ are constants with $\hbar\to0$. I've tried changing co-ordinates to $a_i=\sin(y_i\pi/\ell_p)$, then changing to polar co-ordinates, but I end up with an integral which diverges from the Jacobian $|\partial(y_1, y_2, y_3, y_4)/\partial(a_1, a_2, a_3, a_4)|$ term. Any suggestions?

EDIT: by "an integral which diverges from the Jacobian $|\partial(y_1, y_2, y_3, y_4)/\partial(a_1, a_2, a_3, a_4)|$ term." I mean changing the co-ordinates to $a_i$ where $i=1,2,3,4$ gives us an integral of the form $$\int_{\mathbb{R}}dt\int_0^{1}dy_i\hat{\rho}(t)|J|\mathrm{e}^{\left(\frac{it}{\hbar}\left(E-(\frac{\ell_p}{\pi})\sqrt{a_1^2+a_2^2+a_3^2+a_4^2}\right)\right)}+\int_{\mathbb{R}}dt\int_1^{0}dy_i\hat{\rho}(t)|J|\mathrm{e}^{\left(\frac{it}{\hbar}\left(E-(\frac{\ell_p}{\pi})\sqrt{a_1^2+a_2^2+a_3^2+a_4^2}\right)\right)},$$ and because in the second integral, the Jacobian term has a negative value in the integration domain, this induces a minus sign, therefore $$2\int_{\mathbb{R}}dt\int_0^{1}dy_i\hat{\rho}(t)|J|\mathrm{e}^{\left(\frac{it}{\hbar}\left(E-(\frac{\ell_p}{\pi})\sqrt{a_1^2+a_2^2+a_3^2+a_4^2}\right)\right)}.$$ When I change to polar coordinates, the integral then has the form $$\int_{\mathbb{R}}dt\int_0^{2}dr\int_0^{\pi}d\theta\int_0^{\pi}d\phi\int_0^{2\pi}d\alpha\frac{\mathrm{e}^{\frac{it}{\hbar}(E-\frac{\ell_p}{\pi}r)}\left|r^3\sin^2\theta\sin\phi\right|\hat{\rho}(t)}{\sqrt{1-r^2 \cos ^2(\theta )} \sqrt{1-r^2 \sin ^2(\alpha ) \sin ^2(\theta ) \sin ^2(\phi )}}\cdot \\ \frac{1}{\sqrt{1-r^2 \sin ^2(\theta ) \cos ^2(\phi )} \sqrt{1-r^2 \cos ^2(\alpha ) \sin ^2(\theta ) \sin ^2(\phi )}}$$ where I have left out the "pointless" constants. Then using the method of stationary on the $t$ and the $r$ integral leaves the three angle integrals I've tired picking a value for $E$ and tried solving numerically, this resulting integral diverges. And I don't think you can actually solve this explicitly either.

I have added some new terms and the square root to the integral, because when I wrote the question I was in a hurry and forgot them at first writing, sorry.

• can you please remove useless constants? i get eyecancer from this--- :( – tired Sep 9 '16 at 14:41
• What is the phrase "with an integral which converges from the Jacobian" intended to mean? – Vladimir Sep 9 '16 at 14:45