How to evaluate $\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$

Question

I saw this problem $$\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$$ on my problem book but I have no idea how to evaluate it. I reduced the integral like this

by king's rule $$I_1= \int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)}dx =\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{1+\cos^2(x)}dx $$ by using the double angle rule

$$I_1=\int_0^{\frac{\pi}{2}} \frac{\ln(2\sin(x/2) \cos(x/2))}{2\cos(x/2)^2}dx$$ we let $x/2=x$ then $dx=2dx$ $$I_1=\int_0^{\frac{\pi}{4}} \frac{\ln(2)}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\sin(x))}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$$

now I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(2)}{\cos(x)^2}dx$ by $I_2$ , and I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\sin(x))}{\cos(x)^2}dx$ by $I_3$ , and I will denote $\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$ by $I_4$

$I_2$ is easy to do $$I_3=\frac{\ln(\tan(x))}{\cos(x)^2}dx +I_4$$ I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(x))}{\cos(x)^2}dx$ by $I_5$ which is easy to do.

the problem was to evaluate $2I_4$ which I couldn't do

Answer 1

Alternatively

\begin{align} &\int_0^{\frac{\pi}{2}} \frac{\ln(\cos x)}{1+\sin^2x}dx\\ =& -\int_0^{\frac{\pi}{2}} \int_0^1 \frac{t\sin^2x}{(1+\sin^2x)(1-t^2 \sin^2 x)}dt\ dx\\ =& -\frac\pi{2\sqrt2}\int_0^1 \frac1{\sqrt2+t}dt = {-\frac{\pi}{2\sqrt{2}}\ln\bigg(1+\frac{1}{\sqrt{2}}\bigg)} \end{align}

Answer 2

Your second form of the integral can be rewritten as

$$I = \int_0^{\frac{\pi}{2}}\frac{\sec^2x\cdot\log(\sin x)}{2+\tan^2x}dx = \frac{\log (\sin x)}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\Biggr|_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\frac{\cot x}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\:dx$$

$$= 0 - \int_0^{\frac{\pi}{2}}\frac{\cot x}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\:dx$$

To solve the remaining piece, denote

$$I[a] = \int_0^{\frac{\pi}{2}}\cot x \tan^{-1}(a\tan x)\:dx \implies I'[a] = \int_0^{\frac{\pi}{2}}\frac{dx}{1+a^2\tan^2x}$$

$$ = \int_0^{\frac{\pi}{2}}\frac{\sec^2x\:dx}{(1+a^2\tan^2x)(1+\tan^2x)} = \frac{1}{1-a^2}\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan^2x}-\frac{a^2}{1+a^2\tan^2x}\right)\sec^2x\:dx$$

In other words

$$ I'[a] = \frac{\pi}{2(1+a)} \implies I[a] = \frac{\pi}{2}\log(a+1)+C$$

Plugging in $I[0]=0$ gives $C=0$, thus the integral we want is

$$I = -\frac{1}{\sqrt{2}}I\left[\frac{1}{\sqrt{2}}\right] = \boxed{-\frac{\pi}{2\sqrt{2}}\log\left(1+\frac{1}{\sqrt{2}}\right)}$$

$$$$

Answer 3

Map $x \mapsto \arctan{x}$ so that

$$ \begin{align} I &:= \int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\cos x\right)}{1+\sin^{2}x}dx \\ &= \int_{0}^{\infty}\frac{\ln\left(\cos\left(\arctan x\right)\right)}{1+\sin^{2}\left(\arctan x\right)}d\left(\arctan x\right) \\ &= -\int_{0}^{\infty}\frac{\ln\left(1+x^{2}\right)}{4x^{2}+2}dx \\ &= -\frac{1}{4}\int_{-\infty}^{\infty}\frac{\ln\left(1+x^{2}\right)}{2x^{2}+1}dx. \\ \end{align} $$

Let $f(z)$ be the integrand as a function of $z$. We can write it as $$ \begin{align} f(z) &= \frac{\log\left(z+i\right)+\log\left(z-i\right)}{2z^{2}+1} \\ &= \frac{1}{2z^{2}+1}\cdot\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ \end{align} $$ where we can justify splitting $\log\left(z^2+1\right)$ by defining $\arg(z-i) \in (\pi/2, 5\pi/2)$ and $\arg(z+i) \in (-5\pi/2, -\pi/2)$. We integrate over the following keyhole contour $C$ illustrated below.

This contour has a radius of $R$. The green x symbols are the poles $i/\sqrt{2}$ and $-i/\sqrt{2}$ and the branch cuts are highlighted in orange. And $\Gamma$ is the union of the two outer circular arcs.

Cauchy's Residue Theorem allows us to express $\displaystyle \oint_C f(z)dz$ as

$$ 2\pi i \mathop{\mathrm{Res}}_{z = i/\sqrt{2}} f(z) = \left(\int_{-R}^{R} + \int_{\Gamma} + \int_{\lambda_1} + \int_{\gamma} + \int_{\lambda_2}\right)f(z)dz $$

As $R \to \infty$, the integrals over $\Gamma$ decay to $0$. As the gap between the line segments shrinks, the radius of $\gamma$ also shrinks and forces the integral over $\gamma$ to decay to $0$ as well.

When $\lambda_1,\lambda_2 \to \Lambda$, we get

$$ \begin{align} \lim_{\lambda_1,\lambda_2 \to \Lambda} \left(\int_{\lambda_1} + \int_{\lambda_2}\right)f(z)dz =& \lim_{\lambda_1 \to \Lambda}\int_{\lambda_1}\frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ & + \lim_{\lambda_2 \to \Lambda}\int_{\lambda_2}\frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ =& -\int_{\Lambda} \frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{5\pi}{2}\right)\right) \\ =& + \int_{\Lambda} \frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{\pi}{2}\right)\right) \\ =& -2\pi i \int_{i}^{i \infty}\frac{dz}{2z^2+1} \\ =& -\sqrt{2} \pi \coth^{-1}{\sqrt{2}}. \\ \end{align} $$

Next, we evaluate the residue at the simple pole $i/\sqrt{2}$ of $f(z)dz$ here:

$$ \begin{align} 2\pi i \mathop{\mathrm{Res}}_{z = i/\sqrt{2}} f(z) &= 2\pi i \lim_{z \to i/\sqrt{2}}\frac{z-\frac{i}{\sqrt{2}}}{2z^{2}+1}\cdot\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ &= 2\pi i\frac{-i}{2\sqrt{2}}\left(\ln\left|\frac{i}{\sqrt{2}}+i\right|+\ln\left|\frac{i}{\sqrt{2}}-i\right|+i\arg\left(\frac{i}{\sqrt{2}}+i\right)+i\arg\left(\frac{i}{\sqrt{2}}-i\right)\right) \\ &= 2\pi i\frac{-i}{2\sqrt{2}}\left(\ln\left|\frac{i}{\sqrt{2}}+i\right|+\ln\left|\frac{i}{\sqrt{2}}-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{3\pi}{2}\right)\right) \\ &= \frac{\pi}{\sqrt{2}}\left(\ln\left(1+\frac{1}{\sqrt{2}}\right)+\ln\left(1-\frac{1}{\sqrt{2}}\right)\right). \\ \end{align} $$

Finally, we get $$ \begin{align} I &= -\frac{1}{4}\left(\sqrt{2}\pi\coth^{-1}\left(\sqrt{2}\right)+\frac{\pi}{\sqrt{2}}\left(\ln\left(1+\frac{1}{\sqrt{2}}\right)+\ln\left(1-\frac{1}{\sqrt{2}}\right)\right)\right) \\ &= -\frac{\pi}{2\sqrt{2}}\ln\left(1+\frac{1}{\sqrt{2}}\right) \\ \end{align} $$

and we're done!

Answer 4

With a few substitutions and integration by parts, we find

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \frac{\ln(\cos x)}{1 + \sin^2x} \, dx \\ &= \int_0^1 \frac{\ln y}{(2-y^2) \sqrt{1-y^2}} \, dy & x=\arcsin\sqrt{1-y} \\ &= - \int_0^1 \frac{dy}{(2-y^2) \sqrt{1-y^2}} \left[\int_y^1 \frac{dz}z\right] \\ &= - \int_0^1 \frac{dz}z \left[\int_0^z \frac{dy}{(2-y^2) \sqrt{1-y^2}}\right] \\ &= - \frac1{\sqrt2} \int_0^1 \arctan \left(\frac1{\sqrt2} \frac z{\sqrt{1-z^2}}\right) \, \frac{dz}z \\ &= -\frac1{2\sqrt2} \int_0^\infty \frac{\arctan w}{w\left(\frac12+w^2\right)} \, dw & z=\frac{\sqrt2\,w}{\sqrt{1+2w^2}} \end{align*}$$

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Edit: On second thought, the earlier suggestion to consider the parameterized integral,

$$I(a,b) = \int_0^\infty \frac{\arctan x}{x^a \left(x^2+b^2\right)} \, dx, \qquad (a,b) \in (0,1)^2,$$

by following along with Svyatoslav's approach shown here, won't work because the integrals to either side of the branch cut due to $x^a$ would vanish as $a\to1$.

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Instead, we can take advantage of the integrand's symmetry,

$$I = -\frac1{4\sqrt2} \int_{-\infty}^\infty \frac{\arctan w}{w \left(\frac12+w^2\right)} \, dw$$

and tackle $$\oint_C \frac{\arctan z}{z \left(\frac12+z^2\right)} \, dz = \oint_C \frac{i \ln \left\lvert\frac{1-iz}{1+iz}\right\rvert - \arg \frac{1-iz}{1+iz}}{z \left(1+2z^2\right)} \, dz$$ along a semicircular contour $C$ as shown below. Pink points are poles at $z=\pm \dfrac i{\sqrt2}$ and a removable singularity at $z=0$; blue points are branch points at $z=\pm i$, and the dashed lines indicate where we take the cuts.

At this point we basically reproduce Accelerator's solution with some slight differences peppered in.

Answer 5

Just for fun, let's see what happens if we use the approach that Lars Ahlfors uses to evaluate $\int_{0}^{\pi} \ln(\sin x) \, \mathrm dx$ in the textbook Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable.

Let $\ln$ be the principal branch of the logarithm.

For $- \frac{\pi}{2} < x < \frac{\pi}{2}$, we have the identity $$\Re \ln(1+e^{2ix}) = \ln \left(\sqrt{\left(1+\cos(2x)\right)^{2}+ \sin^{2}(x)} \right) = \frac{1}{2} \, \ln \left(4 \cos^{2}(x) \right) = \ln(2 \cos x).$$

Using this identity, we have $$ \begin{align} \int_{0}^{\pi/2} \frac{\ln \left(\cos x \right)}{1+\sin^{2}(x)} \, \mathrm dx &= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\ln \left(\cos x \right)} {1+\sin^{2}(x)} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{1}{2}\int_{- \pi/2}^{\pi/2} \frac{\ln(2)}{1+\sin^{2}(x)} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln(2)}{1+2t^{2}} \, \mathrm dt \\ &=\frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{\pi \ln(2)}{2 \sqrt{2}}. \end{align}$$

The function $1+e^{2iz}$ is real and nonpositive when $\Re(z) = \pi \, \frac{2n+1}{2}$ ($n \in \mathbb{Z}$) and $\Im(z) \le 0$.

If we omit these half-lines, the principal branch of $\ln(1+e^{2iz})$ is well defined and analytic.

So let's integrate $$f(z) = \frac{\ln \left(1+e^{2iz} \right)}{1+ \sin^{2}(z)} $$ around a tall rectangle contour with vertices at $z= - \pi/2$, $z= \pi/2$, $z= \pi/2 + i R$, $z= -\pi/2 + iR$.

The contour also includes small semicircles about the branch points at $z= \pm \pi/2$, but the integrals along the semicircles vanish as the radii of the semicircles go to zero since $\lim_{z \to \pm \pi/2} z f(z) =0$.

(Near $z= \pm \pi/2$, $\ln(1+e^{2iz}) \sim \ln\left(z \mp \frac{\pi}{2} \right)$.)

The integrals on the vertical sides of the contour are purely imaginary.

And the integral on the top of the contour vanishes as $R \to \infty$ since the magnitude of $f(z)$ decays exponentially to zero.

Since $ \sin(z) =\pm i$ when $z= n \pi \pm i \operatorname{arsinh}(1)$, the only singularity inside the contour is a simple pole at $z= i \operatorname{arsinh}(1) = i\ln (1+ \sqrt{2})$. (See appendix.)

So we have

$$ \begin{align} \Re \int_{-\pi/2}^{\pi/2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx &= \Re \, 2 \pi i \operatorname{Res} \left[f(z) , z=i \ln(1+ \sqrt{2}) \right] \\ &= \Re \, 2 \pi i \lim_{z \to i \ln(1+\sqrt{2})} \frac{\ln \left(1+e^{2iz} \right)}{\sin(2z)} \\ &= \Re \, 2 \pi i \, \frac{\ln \left(1+ \frac{1}{(1+\sqrt{2})^{2}} \right)}{\frac{1}{2i} \left(\frac{1}{(1+\sqrt{2})^{2}} - (1+\sqrt{2})^{2}\right)} \\&= \pi \, \ln \left(\frac{4+2 \sqrt{2}}{3+2 \sqrt{2}}\right) \frac{3+2 \sqrt{2}}{4+3 \sqrt{2}} \\ &= \frac{\pi}{\sqrt{2}} \, \ln (4-2\sqrt{2}). \end{align}$$

And therefore

$$ \begin{align} \int_{0}^{\pi/2} \frac{\ln \left(\cos x \right)}{1+\sin^{2}(x)} \, \mathrm dx &= \frac{1}{2} \left(\frac{\pi }{\sqrt{2}} \, \ln \left(4-2 \sqrt{2}\right) \right) - \frac{\pi \ln(2)}{2 \sqrt{2}} \\ &= \frac{\pi}{2 \sqrt{2}} \, \ln \left(2- \sqrt{2} \right) \\ &= \frac{\pi}{2 \sqrt{2}} \, \ln \left(\frac{2}{2+\sqrt{2}} \right) \\ &= - \frac{\pi}{2 \sqrt{2}} \, \ln \left(1+ \frac{1}{\sqrt{2}} \right). \end{align}$$

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Appendix

$\sin(\pm iz) = \pm i \sinh(z) $

$\sin (n \pi +z) = (-1)^n \sin(z) $

$\sin \left(n \pi \pm i \operatorname{arsinh}(1)\right) = (-1)^n \sin\left( \pm i \operatorname{arsinh}(1) \right) = \pm i (-1)^{n} \sinh\left(\operatorname{arsinh}(1) \right) = \pm i (-1)^{n}$

Answer 6

Let $t=\tan x$, then $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln (\cos x)}{\sec ^2 x+\tan ^2 x} d x \\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \end{aligned} $$ Using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi), $$ and the identity $\ln(a^2+b^2)=2\Re (\ln(a-bi))$, we have $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t&=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \\ & =2 \Re \int_{-\infty}^{\infty} \frac{\ln (1-t i)}{1+2 t^2} d t \\ & =2 \Re \int_\gamma \frac{\ln (1-z i)}{1+2 z^2} d z \\ & =\Re\left[2 \pi i \cdot \lim _{z \rightarrow \frac{i}{\sqrt{2}}}\left(z-\frac{i}{\sqrt{2}}\right) \frac{\ln (1-z i)}{2\left(z^2+\frac{1}{2}\right)}\right] \\ & =\frac{\pi}{\sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) \\ & \end{aligned} $$

Answer 7

Let $t=\tan x$, then $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln (\cos x)}{\sec ^2 x+\tan ^2 x} d x \\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \end{aligned} $$ Considering the parametrised integral $$ I(a)=\int_0^{\infty} \frac{\ln \left(1+a t^2\right)}{1+2 t^2} d t $$ Differentiating $I(a)$ w.r.t $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{t^2}{\left(1+a t^2\right)\left(1+2 t^2\right)} d t \\ & =\frac{1}{2-a} \int_0^{\infty}\left(\frac{1}{1+a t^2}-\frac{1}{1+2 t^2}\right) d t \\ & =\frac{1}{2-a}\left[\frac{1}{\sqrt{a}} \tan ^{-1}(\sqrt{a} t)-\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{2} t\right]_0^{\infty}\\&= \frac{\pi}{2}\left[\frac{1}{(2-a) \sqrt{a}}-\frac{1}{(2-a) \sqrt{2}}\right]\\&= \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{(\sqrt{2}+\sqrt{a}) \sqrt{a}} \end{aligned} $$ Integrating back from $0$ to $1$ gives $$ \begin{aligned} I(1) & = \frac{\pi}{2 \sqrt{2}} \int_0^1 \frac{1}{(\sqrt{2}+\sqrt{a}) \sqrt{a}} d x \\ & = \frac{\pi}{2 \sqrt{2}}[2 \ln (\sqrt{x}+\sqrt{2})]_0^1 \\ & =\frac{\pi}{\sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) \end{aligned} $$ Hence $$ \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t=-\frac{1}{2} I(1)=-\frac{\pi}{2 \sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) $$