What is the integral over a hemisphere for a specific function?

[a_Vatar]

Hi,
I'm trying to evaluate this integral over a hemisphere:
$$\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw$$
where dw - solid angle measure,$$\phi$$ is azimuthal angle and $$\theta$$
is polar angle.
Thus we have:
$$\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw$$ = $$\int \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi$$.
over hemishere.

Integral = $$\int^{2 * pi}_{0} \int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi$$

evaluate inner integral:
making substitution u = cos($$\theta$$),
$$\int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta = <br /> - \int^{pi/2}_{0}u^{a*cos^2(\phi) + b*sin^2(\phi)} du = \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1}$$

That's where I get stuck with integration wrt $$\phi$$
Apparently this integral has to evaluate to $$\frac{1}{\sqrt{(a+1)(b+1)}}$$

 

[AiRAVATA]

Use the substitution

$$t=\tan \frac{\phi}{2},$$

and you end up with a rational integral.

 

[a_Vatar]

$$\int \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1} d\phi$$


after letting $$t=\tan \frac{\phi}{2}$$ I get

$$\int \frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}$$

this does not look to facinating, so I let $$u=t^2$$ and this simplifies it a bit further, but introduces $$\sqrt{u}$$ into denominator;


I tried pluging original $$\int \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1} d\phi$$
in the online integrator here (http://integrals.wolfram.com/index.jsp) and the result is reasonably simple, but when I plugged the "t subsitution integral", the result is way more complex! Btw, the u substitution apprarently yeilds result with complex number, where these come from I have no idea, cause the original integral is simple.

I don't know if Mathemathica actually can give you the intermediate steps of integration e.g. how it got the result, but if it can, could some one having access to it post integration process here please?

 

[AiRAVATA]

$$\int \frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}$$

this does not look to facinating

True! But if you want to solve it with real variable, now you can use partial fractions, i.e.

$$\frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}=\frac{1+t^2}{(\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}) (\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a})}$$

(assuming $a>b$), hence

$$\frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}=\frac{1}{\sqrt{1+a}}\left( \frac{1}{\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}}+ \frac{1}{\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a}}}\right).$$

Using complex variable is fair more simple, just do the substitution $2 \cos \phi=z+1/z,\,2i \sin \phi=z-1/z$ and use Cauchy's Theorem.

P.S. You are missing a factor of two multipliying the integral.

 

[a_Vatar]

How did you get this expression?

$$\frac{1+t^2}{(\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}) (\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a})}$$

when I factor

$${a(1-t^2)^2 + 4bt^2 + (1+t^2)^2} = (a+1)t^4 -2(a-2b-1)t^2 + (a+1)$$
then say,

$$u=t^2$$ and
$$u1,2 = \frac{(a-2b-1)\pm 2\sqrt{(1+b)(b-1)}}{a+1}$$,
thus
$$(a+1)t^4 -2(a-2b-1)t^2 + (a+1) = (t^2 - u1)(t^2 - u2)$$

 

[AiRAVATA]

Simply write the expression

$$(a+1)t^4 -2(a-2b-1)t^2 + (a+1)=(\sqrt{a+1}t^2+\alpha t+\sqrt{a+1})(\sqrt{a+1}t^2+\beta t +\sqrt{a+1})$$

and compute $\alpha,\,\beta$.

 

[a_Vatar]

but where do this coefficient come from:
$$\sqrt{a+1}$$
Is there a general formula to expend a quartic? Or am I missing something obvious
Thanks.

 

[AiRAVATA]

Well, I started trying to express the quartic as a product of two second order polinomials, ending up with five equations for six coefficients, which meant there is no unique expression, so the logical approach was to use the symmetry of the highest and lowest term.