Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers

Question

Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers

I have spent about a day thinking over this problem. I tried integration by parts, differentiating under integral sign (Feynman's trick, with respect to $a, b$), using some trigonometric and logarithmic properties like changing $\cos^2\theta$ to $\cos2\theta$ and hereafter some logarithmic properties, etc., but failed miserably.

Also tried to use the property that $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ in-between, but still to no avail. I also tried to find similar questions on MSE but did not get a related one.

Can someone please help me to solve this integral?

Edit

My try (Feynman's trick) :

Let $$I(a)=\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) d\theta$$

Hence $$I'(a) =\frac 1a \int_0^{\pi/2} \frac {2a^2\cos^2\theta d\theta}{a^2\cos^2\theta +b^2\sin^2\theta}$$ $$=\frac 1a\left[ \frac {\pi}{2}+\int_0^{\pi/2} \frac {a^2\cos^2\theta -b^2\sin^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}\right]$$

Wherein between I broke $2a^2\cos^2\theta=a^2\cos^2\theta +b^2\sin^2\theta+a^2\cos^2\theta -b^2\sin^2\theta$

But now how do I continue further

Answer 1

Let $I(a,b)=\int_0^{\pi/2} \log(a^2\cos^2(\theta)+b^2\sin^2(\theta))\,d\theta$. Differentiating under the integral with respect to $a^2$ reveals

$$\begin{align} \frac{\partial I(a,b)}{\partial (a^2)}&=\int_0^{\pi/2}\frac{1}{a^2+b^2\tan^2(\theta)}\,d\theta\\\\ &=\frac{\pi/2}{a(a+b)}\tag1 \end{align}$$

Integrating $(1)$ with respect to $a^2$, we obtain

$$I(a,b)=\pi \log(a+b)+C$$

For $a=b$, $I(a,a)=\pi \log(a)$ from which we find that $C=-\pi\log(2)$.

Putting it all together yields

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

---

NOTE:

We see from symmetry that $I(a,b)=\pi \log\left(\frac{|a|+|b|}2\right)$ $\forall (a,b)\in \mathbb{R}^2\setminus (0,0)$.

Answer 2

If we assume $a,b>0$ and set $$ I(a,b)=\int_{0}^{\pi/2}\log(a^2\cos^2\theta+b^2\sin^2\theta)\,d\theta $$ we have $I(a,b)=I(b,a)$ from the substitution $\theta\mapsto\frac{\pi}{2}-\theta$. On the other hand $I(a,a)=\pi\log(a)$ is trivial, so $I(a,b)=\pi\log\left(\frac{a+b}{2}\right)$ is a very reasonable conjecture. Indeed, it can be proved by computing $$ \frac{\partial I}{\partial a} = \int_{0}^{\pi/2}\frac{2a\cos^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta $$ as suggested in the comments, i.e. via $\theta\mapsto\arctan u$ and partial fraction decomposition.

An alternative approach is to notice that $$\begin{eqnarray*} I(a,b) &=& 2\,\text{Re}\int_{0}^{\pi/2}\log\left(a\cos\theta+ib\sin\theta\right)\\&=&\pi\log\left(\frac{a+b}{2}\right)+\text{Re}\int_{0}^{\pi}\log\left(e^{i\theta}+\frac{a-b}{a+b}\right)\,d\theta\end{eqnarray*}$$ where the last integral is a purely imaginary number by the residue theorem.

Answer 3

Integrate by parts, then substitute $u=\tan\theta$ :

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \ln\left(a^2\cos^2\theta + b^2\sin^2\theta\right) \, d\theta \\ &= \pi\ln a + \int_0^{\tfrac\pi2} \ln\left(1 + \left(\frac{b^2}{a^2}-1\right) \sin^2\theta\right) \, d\theta \\ &= \pi \ln b - 2\left(\frac{b^2}{a^2}-1\right) \int_0^{\tfrac\pi2} \frac{\theta \sin\theta \cos\theta}{1 + \left(\frac{b^2}{a^2}-1\right) \sin^2\theta} \, d\theta \\ &= \pi \ln b - 2\left(\frac{b^2}{a^2}-1\right) \int_0^\infty \frac{u \arctan u}{\left(u^2+1\right) \left(\frac{b^2}{a^2} u^2 + 1\right)} \, du \end{align*}$$

Now, for some real $\beta\neq0$, let

$$J(\alpha) = \int_0^\infty \frac{u \arctan(\alpha u)}{\left(u^2+1\right) \left(\beta^2u^2+1\right)} \, du \quad [I(0)=0]$$

Differentiate w.r.t. $\alpha$ and evaluate the subsequent integral. Partial fractions are our friend.

$$\frac{\partial J}{\partial \alpha} = \int_0^\infty \frac{u^2}{\left(u^2+1\right) \left(\beta^2u^2+1\right) \left(\alpha^2u^2+1\right)} = \frac\pi{2\left(\beta^2-1\right)} \left[\frac1{\alpha+1} - \frac1{\alpha+\beta}\right]$$

Integrate w.r.t. $\alpha$:

$$J(\alpha) = \frac\pi{2\left(\beta^2-1\right)} \int_0^\alpha \left(\frac1{t+1} - \frac1{t+\beta}\right) \, dt = \frac\pi{2\left(\beta^2-1\right)} \ln \frac{\beta(\alpha+1)}{\alpha+\beta}$$

Finally, replace $\beta=\dfrac ba$ and let $\alpha\to1$ to recover $I$:

$$J(1) = \frac{\pi a^2}{2\left(b^2-a^2\right)} \ln \frac{2b}{a+b} \implies \boxed{I = \pi \ln \frac{a+b}2}$$

Answer 4

Let $r= \frac{a-b}{a+b}$ \begin{align} & \int^{{\pi}/{2}}_{0}\ln{\left(a^2\sin^2t+b^2\cos^2t\right)}dt\\ = & \int^{\pi/2}_{0} \bigg[2\ln\frac{a+b}2+ \ln(1+r^2 -2r\cos 2t)\bigg] dt = \pi\ln\frac{a+b}2 \end{align} where \begin{align} & \int^{{\pi}/{2}}_{0} \ln(1+r^2 -2r\cos 2t) dt\\ = & \int^{\pi/2}_{0} \int_0^{r} \frac{2s-2\cos2t}{1+s^2 -2s\cos 2t}ds \ dt\\ = &\ \int_0^{r} \left(\frac1s\tan^{-1} \frac{s\sin2t}{s\cos 2t-1}\right)_{t=0}^{t=\frac\pi2}ds = \int_0^r 0 \ ds=0 \end{align}