$\int_0^{\frac{\pi}{2}}\ln(\sin^2 x+k^2\cos^2 x)dx$ not by differentiation under the integral?

Question

Show that $$\int_0^{\frac{\pi}{2}}\ln(\sin^2 x+k^2\cos^2 x)dx=\pi\ln \frac{1+k}{2}$$

This is an exercise from Edwards Treatise on Integral Calculus II pg.188.

What solution to this problem can be given ?

In the chapter in Edwards he introduces the technique of substitution $x\mapsto \frac{\pi}{2}-x$. And proves Euler's formulas $$\int_0^{\frac{\pi}{2}}\ln \sin x dx =\int_0^{\frac{\pi}{2}}\ln \cos x dx=-\frac{\pi}{2}\ln 2$$ The problem can be written as

$$\int_0^{\frac{\pi}{2}}\ln(\tan^2 x+k^2 )dx+2\int_0^{\frac{\pi}{2}}\ln(\cos x)dx=\pi\ln(1+k)-\pi\ln 2$$

Which reduces the question to $$\int_0^{\frac{\pi}{2}}\ln(\tan^2 x+k^2 )dx=\pi\ln(1+k)$$

What proof can people give of this equation ?

The only success I have had is to differentiate with respect to $k$, and the result is easy to evaluate. However, it seems that Edwards has at this stage not discussed that method, (although he lists it as a technique, and there is a chapter on it in part I) so is there another solution ? Of course, these old books may be playing by different rules. And the Tripos type questions dont follow any pedagogical pattern. But I guess I really want to know if someone has another solution.

Answer 1

First of all, we can note that $ \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}= \int_{0}^{\frac{\pi}{2}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x} $, $ \ \ \ \ \left(y=\frac{\pi}{2}-x\right)$

Substracting the RHS from the LHS, then multiplying by $ 2 $, gives $ \int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan^{2}{x}\right)}\,\mathrm{d}x} = 0$.

\begin{aligned}\int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan^{2}{x}+k^{2}\right)}\,\mathrm{d}x}&=k^{2}\int_{0}^{\frac{\pi}{2}}{\int_{0}^{1}{\frac{\mathrm{d}y\,\mathrm{d}x}{\tan^{2}{x}+k^{2}y}}}\\ &=k^{2}\int_{0}^{1}{\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x\,\mathrm{d}y}{\tan^{2}{x}+k^{2}y}}}\\ &=k^{2}\int_{0}^{1}{\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{\cos^{2}{x}\left(\tan^{2}{x}+k^{2}{y}\right)\left(1+\tan^{2}{x}\right)}}\,\mathrm{d}y}\\ &=k^{2}\int_{0}^{1}{\int_{0}^{+\infty}{\frac{\mathrm{d}x}{\left(x^{2}+k^{2}y\right)\left(1+x^{2}\right)}}\,\mathrm{d}y}\\ &=k^{2}\int_{0}^{1}{\int_{0}^{+\infty}{\left(\frac{1}{\left(1-k^{2}y\right)\left(x^{2}+k^{2}y\right)}-\frac{1}{\left(1-k^{2}y\right)\left(1+x^{2}\right)}\right)\mathrm{d}x}\,\mathrm{d}y}\\ &=k^{2}\int_{0}^{1}{\left(\frac{1}{1-k^{2}y}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{x^{2}+k^2y}}-\frac{1}{1-k^{2}y}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}}}\right)\mathrm{d}y}\\ &=k^{2}\int_{0}^{1}{\left(\frac{\pi}{2\left(1-k^{2}y\right)}\left(\frac{1}{k\sqrt{y}}-1\right)\right)\mathrm{d}y}\\ &=k\pi\int_{0}^{1}{\frac{\mathrm{d}y}{2\sqrt{y}\left(1+k\sqrt{y}\right)}}\\ &=\pi\int_{0}^{1}{\frac{k}{1+ky}\,\mathrm{d}y}\\ \int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan^{2}{x}+k^{2}\right)}\,\mathrm{d}x}&=\pi\ln{\left(1+k\right)}\end{aligned}

Answer 2

I doubt this was the approach Edwards had in mind, but if we make the substitution $u = \tan x$, the integral becomes $$ \begin{align}\int_{0}^{\pi /2}\ln(\sin^2 x+k^2\cos^2 x) \, \mathrm dx &=\int_{0}^{\infty} \frac{\ln \left(\frac{u^{2}+k^{2}}{u^{2}+1} \right)}{1+u^{2}} \, \mathrm du \\ & = \int_{0}^{\infty} \frac{\ln(u^{2}+k^{2})}{u^{2}+1} \, \mathrm du - \int_{0}^{\infty} \frac{\ln(u^{2}+1)}{u^{2}+1} \, \mathrm du. \end{align}$$

And by integrating the function $$f(z) = \frac{\ln(z+ik)}{z^{2}+1}, \quad k>0,$$ around a semicircular contour in the upper half of the complex plane, we find that $$ \int_{-\infty}^{\infty} \frac{\ln(x+ik)}{x^{2}+1} \, \mathrm dx = 2 \pi i \operatorname{Res}[f(z), i] = \pi \ln (i+ik). $$

(The branch cut is entirely in the lower half-plane if we use the principal branch of the logarithm.)

Equating the real parts on both side of the equation, we get $$\frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln(x^{2}+k^{2})}{x^{2}+1} \, \mathrm dx = \pi \ln (1+k).$$

Therefore, $$\int_{0}^{\pi /2}\ln(\sin^2 x+k^2\cos^2 x) \, \mathrm dx = \pi \ln (1+k) - \pi \ln 2 = \pi \ln \left( \frac{1+k}{2} \right).$$

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A generalization:

Assume that $k \ge 0$ and $ 0 < \theta < \pi$.

Let $$I(k, \theta) = \int_{-\pi/2}^{\pi/2} \ln \left(\sin^{2}(x)+k \sin(2x) \cos (\theta) +k^{2} \cos^{2}(x)\right) \, \mathrm dx .$$

Then again making substitution $u = \tan x$, we get

$$\begin{align} I(k, \theta) &= \int_{-\infty}^{\infty} \frac{\ln \left(\frac{u^{2}+2ku \cos (\theta) +k^{2}}{u^{2}+1} \right)}{u^{2}+1} \, \mathrm du \\ &= \int_{-\infty}^{\infty} \frac{\ln(u^{2}+2ku \cos (\theta) + k^{2})}{u^{2}+1} \, \mathrm du - 2 \pi \ln 2 \\ &= \pi \ln (1+2k \sin (\theta) +k^{2}) - 2 \pi \ln 2 \\ &= \pi \ln \left( \frac{1+2k \sin (\theta) +k^{2}}{4} \right). \end{align}$$

An evaluation of the integral on the second line using contour integration can be found here.

Answer 3

There is a quite low-level elementary solution. Using some trigonometric identities and change of variable, we obtain \begin{multline*} \int\limits_0^{\frac \pi 2} \ln(\sin^2 x + k^2\cos^2x)\,dx = \int\limits_0^{\frac \pi 2} \ln\Big(\frac{1-\cos 2x}2+ k^2\frac{1+\cos 2x}2\Big)\,dx = \\ \frac 12 \int\limits_0^\pi \ln\Big(\frac{1-\cos t}2+ k^2\frac{1+\cos t}2\Big)\,dt = \frac 14 \int\limits_0^{2\pi} \ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos t\Big)\,dt. \end{multline*}

Suppose that $0 < k < 1$. Then we have an integral of a continuous function, which equals to the limit of its Riemann sums: \begin{multline*} \frac 14 \int\limits_0^{2\pi} \ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos t\Big)\,dt = \frac \pi{2n} \lim\limits_{n\to\infty} \sum\limits_{j=0}^{n-1}\ln\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos \frac{2\pi j}n\Big) =\\ \frac \pi{2n} \lim\limits_{n\to\infty} \ln \prod\limits_{j=0}^{n-1}\Big(\frac{1+k^2}2-\frac{1-k^2 }2\cos \frac{2\pi j}n\Big) = \frac \pi{2n} \lim\limits_{n\to\infty} \ln \prod\limits_{j=0}^{n-1}\Big(a^2 - 2ab\cos \frac{2\pi j}n + b^2\Big), \end{multline*} where $a = \frac{1+k}{2}$, $b = \frac{1-k}{2}$. Applying the algebraic identity $$ \prod\limits_{j=0}^{n-1}\Big(a^2 - 2ab\cos \frac{2\pi j}n + b^2\Big) = (a^n - b^n)^2, $$ we finally get $$ \frac \pi{2n} \lim\limits_{n\to\infty} \ln (a^n - b^n)^2 = \frac \pi{n} \lim\limits_{n\to\infty} \ln \Big(\Big(\frac{1+k}2\Big)^n - \Big(\frac{1-k}2\Big)^n\Big) = \pi \ln \frac{1+k}2. $$