Q:

trying to find initial condition for an integral depending a variable

Let 
$$ F( \alpha) = \int\limits_0^{\pi/2} \ln(\alpha^2 - \sin^2 x) \mathrm{d}x \; \; \; \; \; \; \alpha>1 $$
let $G(\alpha,x) = \ln(\alpha^2 - \sin^2 x)$ I know $F'( \alpha) = \int\limits_0^{\pi/2} \frac{2 \alpha }{\alpha^2 - \sin^2 x } dx$
Using the substitution $u = \tan x$, I was able to solve this integral and I obtained 
$$ F'(\alpha) = \frac{ \pi }{\sqrt{ \alpha^2 - 1 } } \implies F(\alpha) = \pi \ln \left(\alpha + \sqrt{ \alpha^2 - 1 } \right)+C$$
However, I need to find the value of $C$. How can I do that? I mean, I know I need to find some $b$ constant so that $F(b) = constant$ in the original $F$ from the first equation above. But, I unable to find such $b$. Any help would be greatly appreciated.

A:

I would set the constant by looking at the dependence for $\alpha \to \infty$. Taylor expansion for small $\sin x \ll \alpha$ leads to the result 
$$F(\alpha)= \int_0^{\pi/2}\ln(\alpha^2 -\sin^2 x)  =\frac{\pi}{2} \ln \alpha^2  +O(\alpha^{-2})= \pi \ln \alpha + O(\alpha^{-2}).$$
From your expression, we obtain
$$F(\alpha) = \pi \ln(2 \alpha) +C +o(1)= \pi \ln \alpha + \pi \ln 2 +C +o(1)$$
for the same limit ($\alpha \to \infty$). This sets the constant to $$C= -\pi \ln2.$$

