How to approach solving $\int_0^{\pi/2} \ln(a^2 \cos^2 x +b^2 \sin^2 x ) dx$

Question

I have been attempting to solve this integral for several days in preparation for an exam and keep reaching dead ends.

I(a,b) = $\int_0^{\pi/2} \ln(a^{2}\cos^2(x)+b^{2}\sin^2(x)) dx$, where $a,b>0$

My best approach so far is defining:

$F(x) = I(x, b)$

Then, I know:

$F(b) = I(b, b) = \pi/2 * \ln(b^2)$

Then I am now attempting to find $F'(x)$ using differentiation under the integral sign, and integrating from b to a.

This is where I am stuck - no matter what I do, I cannot solve the integral. Am I in the right direction or should this be solved in a different manner? I also considered something elliptical due to the acos, bsin in the original integral, yet it has also reached a dead end.

Answer 1

Maybe someone is interested in a more "outside the box approach"

Let us define

$$F(a,b)=\int_0^{\pi/2}\log(a^2\cos(x)^2+b^2\sin(x)^2)dx$$

Now

$$ \partial_aF(a,b)=2a\int_0^{\pi/2}\frac{\cos(x)^2}{a^2\cos(x)^2+b^2\sin(x)^2}dx $$

$$ \partial_bF(a,b)=2b\int_0^{\pi/2}\frac{\sin(x)^2}{a^2\cos(x)^2+b^2\sin(x)^2}dx $$

which leads immediatly to the follwoing partial differential equation

$$ a \partial_aF(a,b)+b\partial_bF(a,b)=\pi $$

by seperation of variables the homogenous solution to the above problem is given by $F_h(a,b)=C\frac{b}{a}$ where $C$ is some constant.

By inspection it is also straightforward to see that the particular solution with the required symmetry $a \leftrightarrow b$ is given by $F_p(a,b)=\pi \log(D(a+b))$. Here $D$ is another free parameter which appears through the invariance under scaling of our DE.

The constants of integration might be fixed by the requirement $F(a,a)=\pi \log(a)$ which induces $C=0$ and $D=1/2$. It follows that

$$ F(a,b)=F_h(a,b)+F_p(a,b)=\pi\log\left(\frac{a+b}{2}\right) $$

Edit: I'm totally aware that (also on this site) there are much simpler approaches to such integrals, i was just in the mood to try something different

Answer 2

The integral can be deduced from the Bronstein integral \begin{equation*} \int_{0}^{\pi}\ln(a^{2}+b^{2}-2ab\cos x)\, dx = 2\pi\ln(\max\{a,b\}) \end{equation*} where $0<a,b \in\mathbb{R}$.

See Bronstein Integral 21.42

Assume that $b>a$ (otherwise change $x$ to $\dfrac{\pi}{2}-x$). Then we get \begin{gather*} \int_{0}^{\pi/2}\ln(a^{2}\cos^{2}x+b^{2}\sin^{2}x)\, dx = \int_{0}^{\pi/2}\ln\left(\dfrac{b^{2}+a^{2}}{2}-\dfrac{b^{2}-a^{2}}{2}\cos(2x)\right)\, dx =\\[2ex] \dfrac{1}{2}\int_{0}^{\pi}\ln\left(\left(\dfrac{b+a}{2}\right)^{2}+\left(\dfrac{b-a}{2}\right)^{2}-2\dfrac{b+a}{2}\dfrac{b-a}{2}\cos x\right)\, dx = \pi\ln\left(\max\left\{\dfrac{b+a}{2},\dfrac{b-a}{2}\right\}\right) = \\[2ex]\pi\ln\left(\dfrac{a+b}{2}\right). \end{gather*}

Answer 3

Through the substitutions $x=\arctan(t)$ and $t\leftrightarrow\frac{1}{t}$we have $$\begin{eqnarray*} I(a,b) &=& \int_{0}^{+\infty}\frac{\log(b^2+a^2 t^2)-\log(1+t^2)}{1+t^2}\,dt\\ &=&\int_{0}^{+\infty}\frac{\log(a^2+b^2 t^2)-\log(1+t^2)}{1+t^2}\,dt=I(b,a)\end{eqnarray*}$$

where $$ \frac{\partial}{\partial b} I(a,b) = \int_{0}^{+\infty}\frac{2 b}{\left(1+t^2\right) \left(b^2+a^2 t^2\right)}\,dt =\frac{\pi}{a+b}=\frac{\partial}{\partial a} I(a,b). $$ It follows that $I(a,b) = C+\pi\log(a+b) $, and by setting $a=b=1$ it is trivial that such a constant $C$ has to be $-\pi\log(2)$, since $I(1,1)=0$.

Answer 4

I would use the substitution $\sin(x)=u$. Then we get

$$I(a,b) = \int_0^{\pi/2} \ln(a^2 \, \cos^2(x)+b^2 \, \sin^2(x)) dx = \int_0^{1} \frac{\ln(a^2 \, (1-u^2)+b^2 \, u^2)}{\sqrt{1-u^2}} \, du$$

Then using integration by parts with

$$\frac{d}{dx}(\ln(a^2 \, (1-u^2)+b^2 \, u^2))= \frac{2\,(a^2-b^2)\,u}{b^2+(a^2-b^2)\,u^2}$$ $$\int\frac{1}{\sqrt{1-u^2}}\,du=\arcsin(u)$$

$$I(a,b) = \frac{1}{2}\pi\ln(b^2) - \int_0^{1} \frac{2\,(a^2-b^2)\,u}{b^2+(a^2-b^2)\,u^2} \, \arcsin(u) \, du = \\ \frac{1}{2}\pi\ln(b^2) - \int_0^{1} \frac{2\,\arcsin(u)}{u}\,du + \int_0^{1} \frac{2\,b^2}{u\,(b^2+(a^2-b^2)\,u^2)} \, \arcsin(u) \, du$$

According to Mathematica, the first integral gives

$$\int_0^{1} \frac{2\,\arcsin(u)}{u}\,du = \pi \ln(2)$$

The second integral can also be evaluated with Mathematica, but it gives me quite the cumbersome result. Maybe there is a simpler way to do this.