Evaluating $\int_{0}^{\pi/2}\log\left(a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$

Question

I am trying to evaluate the integral below by differentiating through the integral. Let

$ F(a,b) :=\displaystyle\int_{0}^{\pi/2}\log\left(a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$

For fixed $b$, and letting $g(t) = F(t,b)$ I am trying to justify

$g'(t) = \displaystyle\int_{0}^{\pi/2} \dfrac{2t\cos^2(x)}{t^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)}\,{\rm d}x$

In order to justify this, I need to show

1. $f(t,b) :=\log\left(t^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)\right)\,{\rm d}x$ is integrable over $[0,\pi/2]$
2. $\dfrac{\partial f}{\partial t}$ exists for all $t,x$
3. There exists a dominating function such that $\dfrac{\partial f}{\partial t} \leq g$, a.e and $g \in L^1([0,\pi/2])$

I'm thinking 1) is straightforward since it's continuous, hence measurable, and since it's on a finite interval we can conclude that it is also integrable.

$\dfrac{\partial f}{\partial t} = \dfrac{2t\cos^2(x)}{a^2\cos^2\left(x\right)+b^2\sin^2\left(x\right)}$ so 2) is satisfied

In regards to finding a dominating function, can MVT be used here?

I'm also having some trouble evaluating the integral $g'(t)$, but $F(a,b) = \pi \log \left(\dfrac{a+b}{2}\right) \ \ (a,b>0)$ should be the solution.

Answer 1

Note that: $${a}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right) =\frac{1}{4} \left( 1+{\frac { \left( a-b \right) {{\rm e}^{-2\,ix}}}{a+b}} \right) \left( {\frac {{{\rm e}^{2 \,ix}} \left( a-b \right) }{a+b}}+1 \right) \left( a+b \right) ^{2}$$ $$\left|{\frac { \left( a-b\right)}{a+b}}\right|\le 1,\quad a,b>0$$ then from: $$-\sum _{n=1}^{\infty }{\frac {{r}^{n}{{\rm e}^{inx}}}{n}}=\ln \left( 1-r{{\rm e}^{ix}} \right) $$ show that: $$2\ln \left(\frac{a+b}{2} \right) -2\sum _{n=1}^{\infty } \frac{\left( -{\frac {a-b}{a+b}} \right) ^{n}\cos \left( 2\,nx \right)}{n} = \ln \left({a}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right)\right) $$ and note that: $$\int _{0}^{1/2\,\pi }\!\cos \left( 2\,nx \right) {dx}=0,\quad n\ge 1$$ $$\int _{0}^{1/2\,\pi }\!2\ln \left( \frac{a+b}{2} \right) {dx}=\ln \left(\frac{a+b}{2} \right) \pi $$ and the integration result follows. Furthermore:

$${\frac {\partial }{\partial t}}\ln \left({t}^{2} \cos^{2} \left( x \right) +{b}^{2} \sin^{2} \left( x \right)\right) ={\frac {2t}{{t}^{2}+{b}^{2} \tan^{2} \left( x \right)}}<\frac{2}{t}$$

Answer 2

$$I(a)=\int_0^\frac\pi2\log\bigg(a^2\cos^2x+b^2\sin^2x\bigg)dx\quad=>\quad I'(a)=\int_0^\frac\pi2\frac{2a\cdot\cos^2x}{a^2\cos^2x+b^2\sin^2x}dx$$

$$J(b)=\int_0^\frac\pi2\log\bigg(a^2\cos^2x+b^2\sin^2x\bigg)dx\quad=>\quad J'(b)=\int_0^\frac\pi2\frac{2b\cdot\sin^2x}{a^2\cos^2x+b^2\sin^2x}dx$$

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$$I'(a)=2a\underbrace{\int_0^\frac\pi2\frac{dx}{a^2+b^2\tan^2x}dx}_{t=\tan x}=2a\,\bigg[~\frac{x-\dfrac ba\arctan\bigg(\dfrac ba\tan x\bigg)}{a^2-b^2}~\bigg]_0^\frac\pi2=\frac\pi{a+b}$$

$$J'(b)=2b\underbrace{\int_0^\frac\pi2\frac{dx}{a^2\cot^2x+b^2}dx}_{t=\cot x}=2b\,\bigg[~\frac{\dfrac\pi2-x-\dfrac ab\arctan\bigg(\dfrac ab\cot x\bigg)}{a^2-b^2}~\bigg]_0^\frac\pi2=\frac\pi{a+b}$$

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$\begin{align}F(a,b)=I(a)=\pi\ln(a+b)+C_b\\\\F(a,b)=J(b)=\pi\ln(a+b)+C_a\end{align}=>$ $F(1,1)=0=>C_a=C_b=-\pi\ln2=>F=\pi\,\ln\dfrac{a+b}2$

Answer 3

$\newcommand{\Log}{\operatorname{Log}}$To change the game a bit. First set $x=\arctan(t)$ then we get: \begin{align} F(a,b)&=\int^\infty_0 \frac{\ln\left(\frac{1}{1+t^2}a^2+\frac{t^2}{1+t^2}b^2\right)}{1+t^2}\,dt\\ &=\int^\infty_0 \frac{\ln\left(a^2+b^2t^2\right)}{1+t^2}\,dt-\int^\infty_0 \frac{\ln\left(1+t^2\right)}{1+t^2}\,dt\\ &=\frac{1}{2}\int^\infty_{-\infty} \frac{\ln\left(a^2+b^2t^2\right)}{1+t^2}\,dt-\frac{1}{2}\int^\infty_{-\infty} \frac{\ln\left(1+t^2\right)}{1+t^2}\,dt \end{align} So it is enough to find $I(a,b)$ defined by: \begin{align} I(a,b):=\frac{1}{2}\int^\infty_{-\infty} \frac{\ln\left(a^2+b^2t^2\right)}{1+t^2}\,dt \end{align} for $a,b>0$.

We use a complex analysis method and consider an integral that will be strange in the first place but later it will become clear. So consider \begin{align}\tag{1} \oint_C \frac{\Log(ia+bz)}{1+z^2}\,dz =\int_{C_R}\frac{\Log(ia+bz)}{1+z^2}\,dz+ \int^R_{-R}\frac{\Log(ai+bz)}{1+z^2}\,dz \end{align} where $C$ is a semi circle contour in the upper half plane; $C_R$ is the circle part and the other one is obvious. Moreover $\Log(\cdot)$ is the Principal Log (with this choice we don't have to deal with the branch cut). The only residue inclosed is $z=i$ and therefore by the Residue Theorem we get: \begin{align} \int_{C_R}\frac{\Log(ia+bz)}{1+z^2}\,dz+ \int^R_{-R}\frac{\Log(ai+bz)}{1+z^2}\,dz=2\pi i \text{Res}_{z=i} \frac{\Log(ia+zb)}{1+z^2} \end{align} The integral on $C_R$ vanshies when $R\to\infty$. Now lets have a look at the real part of the one on $[-R,R]$: \begin{align} \Re\left(\int^R_{-R}\frac{\Log(ia+bz)}{1+z^2}\,dz\right) &= \frac{1}{2}\int^R_{-R} \frac{\ln(a^2+b^2t^2)}{1+t^2}\,dt \end{align} Now the choice of $(1)$ becomes clear. Let $R\to \infty$ and we get: \begin{align} I(a,b) = \Re\left(2\pi i \text{Res}_{z=i} \frac{\Log(ia+bz)}{1+z^2}\right) \end{align} Doing the residue yields: \begin{align} I(a,b)=\pi\ln(a+b) \end{align} We also know that $F(a,b)=I(a,b)-I(1,1)=\pi\ln(a+b)-\pi\ln(2)$ so finally we conclude:

$$F(a,b)=\int^{\pi/2}_0\ln(a^2\cos^2(x)+b^2\sin^2(x))\,dx=\pi\ln\left(\frac{a+b}{2}\right)$$ with $a,b>0$.