Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$?

Question

For $a,b>0$, show that

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$ I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.

For $a>b$ then, we can write the $$ a^2\cos^2x +b^2\sin^2x =b^2\cos^2x+b^2\sin^2x + k\cos^2x = b^2 +k\cos^2x $$ where $ k= a^2-b^2$ and similarly, for $ b>a$ we can write $a^2\cos^2x +b^2\sin^2x = a^2+l\sin^2x$ where $l= b^2-a^2$. Hence we have $$ \int_0^{\frac{\pi}{2}}\ln(b^2+k\cos^2x )dx=\\ \int_0^{\frac{\pi}{2}}\left(\ln(b^2) + \ln\left(1+\frac{k}{b^2}\cos^2x\right)\right)dx\cdots(1)$$ and $$\int_0^{\frac{\pi}{2}}\ln(a^2+l\sin^2x )dx \\= \int_0^{\frac{\pi}{2}}\left(\ln(a^2) + \ln\left(1+\frac{l}{a^2}\sin^2x\right)\right)dx, \; \; b>a$$ Since $\left| \frac{k}{b^2}\cos^2 x\right|\leq 1$ so we use the Machlaurin series of $\ln(1+x)$ for $|x|<1$ giving us $$ \pi \ln(b)+\sum_{m\geq 1} \frac{(-1)^{m-1}}{m}\left(\frac{k}{4b^2}\right)^m\int_0^{\frac{\pi}{2}}\cos^{2m}xdx$$ The latter integral is well know result know as Wallis integral thus gives us $$ \frac{\pi}{2}\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\cdots(2)$$ Recalling the generating function of Central binomial coefficients for $|y| <\frac{1}{4}$, ie $$\sum_{m\geq 1} {2m\choose m} y^{m}=\frac{1}{\sqrt{1-4y}}-1.$$ Dividing by $y$ and an integrating from $0$ to $z$ we have $$ \sum_{m\geq 1}\frac{1}{m}{2m\choose m} z^{m} =-2\ln\left(\frac{\sqrt{1-4z}+1}{2}\right) $$ multiplying thoroughly by $-1$ and set $z=-\frac{k}{4b^2}=-\frac{a^2-b^2}{4b^2}$ gives us $$\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\\ = 2\ln\left(\frac{{a+b}}{2}\right)-2\ln b $$ From $(1)$ and $(2)$ we have $$\pi\ln(b)+\pi\ln\left(\frac{a+b}{2}\right)-\pi\ln b \\= \pi\ln\left(\frac{a+b}{2}\right)$$ Similarly, for the case of $b>a$ we replace $ \cos^2x$ by $\sin^2x$, $k$ by $ l$. We obtained the same desired result. Moreover, we can also have the following result.

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\frac{1}{2}\int_0^{\pi}\ln(a^2\cos^2x+b^2\sin^2x)dx=\pi\ln\left(\frac{a+b}{2}\right)$$

Now I'm wishing to know other different approaches/proofs for the aforementioned integral.

Thank you.

Answer 1

Let

$$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln(a^2 \cos^2 x+b^2 \sin^2 x) dx $$

Thus

$$ I(b)=\int_{0}^{\frac{\pi}{2}} \ln(b^2 \cos^2 x+b^2 \sin^2 x) dx=\pi \ln b $$

\begin{align} I'(a)&=\int_{0}^{\frac{\pi}{2}} \frac{2a\cos^2 x}{a^2 \cos^2 x+b^2 \sin^2 x}dx \\ &=2a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{(a^2+b^2 \tan^2 x)(1+\tan^2 x)} \\ &=\frac{\pi}{a+b} \\ \end{align}

Thus

$$ \int_b^a I'(a) da=\int_b^a \frac{\pi}{a+b} da $$

$$ I(a)-I(b)=\pi (\ln(a+b)-\ln(2b)) $$

$$ I(a)=\pi \ln \frac{a+b}{2} $$

Answer 2

I thought it might be instructive to present an approach that relies on contour integration in the complex plane. To that end, we proceed.

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MODIFYING THE INTEGRAL

Let $I(a,b)$ be defined by the integral

$$I(a,b)=\int_0^{\pi/2}\log(a^2\cos^2(x)+b^2\sin^2(x))\,dx\tag1$$

Using the double angle formulas, $\sin^2(x)=\frac{1-\cos(2x)}{2}$ and $\cos^2(x)=\frac{1+\cos(2x)}{2}$, in $(1)$ we obtain

$$\begin{align} I(a,b)&=\int_0^{\pi/2}\log\left(\frac{a^2+b^2}2+\frac{a^2-b^2}{2}\cos(2x)\right)\,dx\tag2\\\\ &=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac14 \int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx\tag3 \end{align}$$

In going from $(2)$ to $(3)$ we enforced the substitution $x\mapsto x/2$ and exploited the evenness of the integrand.

From here, we could use Feynman's trick, differentiate under the integral (tangent half-angle or contour integration), evaluate the integral of the derivative, and finish by integrating. Instead, we will apply contour integration to directly evaluate the integral.

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MOVING TO THE COMPLEX PLANE

Next, we transform the integration on the right-hand side of $(3)$ to the complex plane through the substitution $z=e^{ix}$ to obtain
$$\begin{align} \int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx&=\oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz\tag4 \end{align}$$

where $-1<\alpha =\frac{a^2-b^2}{a^2+b^2}<1$.

We select the principal branch of the logarithm with branch cuts emanating from branch points, extending along the negative real axis, and terminating at $z=-\infty$.

Denote the roots of $z^2+2z/\alpha+1=0$ by $r_1$ and $r_2$.

We will assume that $a<b$ so that $-1<r_1=\frac{a-b}{a+b}<0$ lies inside the unit circle $|z|=1$ while $r_2=\frac{a+b}{a-b}<-1$ lies outside the unit circle.

Then, we have from $(4)$

$$\begin{align} \oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz&=\oint_{|z|=1} \frac{\log\left(-\alpha/2\right)}{iz}\,dz-\oint_{|z|=1} \frac{\log\left(z\right)}{iz}\,dz\\\\ &+\oint_{|z|=1} \frac{\log\left(z-r_1\right)}{iz}\,dz+\oint_{|z|=1} \frac{\log\left(z-r_2\right)}{iz}\,dz\tag5 \end{align}$$

where the negative sign on $\alpha$ is required to ensure that the sum of the arguments of the logarithms is $0$ on the unit circle.

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EVALUATING THE INTEGRALS IN $(5)$

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From the residue theorem, we see that

$$\oint_{|z|=1} \frac{\log\left(\alpha/2\right)}{iz}\,dz=2\pi \log\left(\alpha/2\right)\tag6$$

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From Cauchy's Integral Theorem, we have

$$\begin{align}\oint_{|z|=1}\frac{\log(z)}{z}\,dz&=\lim_{\varepsilon\to0}\left(\int_{-1}^{-\varepsilon}\frac{(\log(-x)-i\pi)-(\log(-x)+i\pi)}{ix}\,dx+\int_{-\pi}^\pi \frac{\log(\varepsilon e^{i\phi})}{i\varepsilon e^{i\phi}}\,d\phi\right)\\\\ &=0\tag7 \end{align}$$

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From the residue theorem, we find that

$$\begin{align} \oint_{|z|=1}\frac{\log(z-r_2)}{iz}\,dz&=2\pi \log(-r_2)\tag8 \end{align}$$

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Finally, from Cauchy's Integral Theorem, we have

$$\begin{align} \oint_{|z|=1}\frac{\log(z-r_1)}{iz}\,dz&=2\pi \log(-r_1)\\\\ &+\lim_{\varepsilon\to0}\left(\int_{-1}^{r_1-\varepsilon}\frac{(\log(-(x-r_1)) -i\pi)-(\log(-(x-r_1))+i\pi)}{ix}\,dx\\\\+\int_{-\pi}^{\pi} \frac{\log(\varepsilon e^{i\phi})}{i(r_1+\varepsilon e^{i\phi})}\,i\varepsilon e^{i\phi}\,d\phi\right)\\\\ &=0\tag9 \end{align}$$

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PUTTING IT ALL TOGETHER

Using $(6)-(9)$, we find that

$$\begin{align} I(a,b)&=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac\pi2 \log(-\alpha/2)+\frac\pi2\log(-r_2)\\\\ &=\pi\log\left(\frac{a+b}{2}\right) \end{align}$$

which agrees with the expected result!