I am stuck at one point and will appreciate if the community can guide me. I have a final expression as part my research which looks something like following:
$$I = \frac{X}{G(\gamma)}$$
where $G(\gamma) = 1+d\cos(\gamma)$ and $\gamma$ is the angle with which the Gain $G$ is defined and $d$ is the directivity.
Similarly, $X = \pi (\epsilon r)^{-\alpha}[\lambda_1(P_1)^{2/\alpha} + \kappa\lambda_2(P_2)^{2/\alpha}]$.
I want to take the average of Gain over angles $\gamma \in [-\pi/2, \pi/2]$, I get $\int_{-\pi/2}^{\pi/2} \frac{G}{\pi} = 1+\frac{2d}{\pi}$. However, I am not sure if I can directly do this for $I$. Which one should I do
$I' = \frac{X}{\pi}\int_{-\pi/2}^{\pi/2}\frac{1}{1+d\cos(\gamma)} d\gamma$
$I' = \frac{\pi X}{\int_{-\pi/2}^{\pi/2} [1+d\cos(\gamma) ]d\gamma}$
$$I=\dfrac{X}{G(\gamma)}$$
If you want to take the average of Gain over angles $\gamma \in [\dfrac{-\pi}{2},\dfrac{\pi}{2}]$,then
$$\dfrac{1}{\pi}\int_{-\pi/2}^{\pi/2}G.d\gamma =\dfrac{1}{\pi}\int_{-\pi/2}^{\pi/2}[1+dcos(\gamma)]d\gamma=1+\dfrac{2d}{\pi} $$
If you want to take the average of $I$ over angles $\gamma \in [\dfrac{-\pi}{2},\dfrac{\pi}{2}]$,then
$$\dfrac{1}{\pi}\int_{-\pi/2}^{\pi/2}I.d\gamma =\dfrac{1}{\pi}\int_{-\pi/2}^{\pi/2}\dfrac{X}{1+dcos(\gamma)}d\gamma =\dfrac{2}{\pi}\int_{0}^{\pi/2}\dfrac{X}{1+dcos(\gamma)}d\gamma $$
Substituting $cos(\gamma)=\dfrac{1-tan^2(\gamma /2)}{1+tan^2(\gamma /2)}$, and taking $tan(\gamma /2)=t$
$$\dfrac{2}{\pi}\int_{0}^{\pi/2}\dfrac{X}{1+d\dfrac{1-tan^2(\gamma /2)}{1+tan^2(\gamma /2)}}d\gamma=\dfrac{2}{\pi}\int_{0}^{1}\dfrac{2X}{(1+d)+(1-d)t^2}dt$$$$=\dfrac{4X}{\pi\sqrt{(1+d)(1-d)}} \arctan(\sqrt{\dfrac{1-d}{1+d}}) $$
If you want to calculate $I$ with the averaged value of Gain, then
$$I=\dfrac{X}{1+\dfrac{2d}{\pi}}$$
EDIT:
$d$ is antenna directivity, and is always $\geq 1$.
$$=\dfrac{2}{\pi}\int_{0}^{1}\dfrac{2X}{(1+d)+(1-d)t^2}dt=\dfrac{2}{\pi}\int_{0}^{1}\dfrac{2X}{(d+1)-(d-1)t^2}dt$$.$$=\dfrac{2}{\pi(d-1)}\int_{0}^{1}\dfrac{2X}{\dfrac{d+1}{d-1}-t^2}dt=\dfrac{2X}{\pi\sqrt{(d+1)(d-1)}}ln\dfrac{\sqrt{d+1}+\sqrt{d-1}}{\sqrt{d+1}-\sqrt{d-1}}$$