Evaluating $\int_{0}^{\frac{\pi}{2}}d\phi \frac{e^{-D \cos^2 \phi}}{A+B\tan^2 \phi}$

Question

In this question, I was looking for a nice closed form for the following integral:

$$\int_{0}^{\pi/2} \frac{e^{-D \cos^2 \phi}}{A+B\tan^2 \phi}d\phi,$$

where $A,B$ and $D$ are constants. As you can see in my answer below, I got it!

Answer 1

My answer:

Let $P=-B$. Thus the original integral takes the form

$$\int_{0}^{\frac{\pi}{2}}d\phi \frac{e^{-D \cos^2 \phi}}{A-P\tan^2 \phi}$$

The change of variable $\cos^2 \phi=x$ give us

$$\frac{1}{2P}\int_{0}^{1} dx \ x^{1/2}(1-x)^{-1/2}(1-\beta x)^{-1}e^{-D x}, $$

where $\beta=(P-A)/P$. Now, using the $3.385^{11}$ formula of Gradstein and Ryzhik's tables of integrals we get

$$\int_{0}^{\frac{\pi}{2}}d\phi \frac{e^{-D \cos^2 \phi}}{A-P\tan^2 \phi}=\frac{\pi}{2P}\Phi_1\left(\frac{1}{2},1,2,-D,\beta \right), $$ where $ \Phi_1\left(\alpha,\beta,\gamma,x,y \right)$ is a confluent hypergeometric series of two variables gives by $9.261.1.^6$.

Answer 2

I don't know if there is a closed form in general. For convenience write $A = B+C$. The change of variable $\cos(\phi)=t$ gives us

$$ \int_0^1 \frac{\exp(-D t^2) t^2}{\sqrt{1-t^2} (C t^2 + B)}\; dt $$

If $|C| < |B|$, this can be expanded in a series in powers of $C$:

$$ \sum_{n=0}^\infty \frac{(-C)^n}{B^{n+1}} \int_0^1 \frac{\exp(-Dt^2) t^{2n+2}}{\sqrt{1-t^2}}\; dt $$

Now if $$J(n) = \int_0^1 \frac{\exp(-Dt^2) t^{2n}}{\sqrt{1-t^2}}\; dt$$ we have $J(0) = \frac{\pi}{2} e^{-D/2} I_0(d/2)$ (where $I_0$ is a modified Bessel function), $J(1) = \frac{\pi}{4} e^{-D/2} (I_0(d/2) - I_1(d/2))$, and, it seems, $$ J(n+2) = -\frac{2n+1}{2D} J(n) + \frac{D + n+1}{D} J(n+1) $$