Any cool ways to solve $\int_{0}^{\frac{\pi}{2}}\log(1+4\sin^2x)\,dx$

Question

As per the title, I have solved the following integral

$$\int_{0}^{\frac{\pi}{2}}\log(1+4\sin^2x)\,dx$$

I would love to see any insights, and solution processes anyone may have in solving it as well.

For anyone who may be interested here is my solution process.

$$I=\int_{0}^{\frac{\pi}{2}}\log(1+4\sin^2(x))\,dx$$

$$I=\int_{0}^{\frac{\pi}{2}}\log((1+4\sin^2(x))\sec^2(x)\cos^2(x))\,dx$$

$$=\int_{0}^{\frac{\pi}{2}}\log(1+5\tan^2(x))\,dx+2\int_{0}^{\frac{\pi}{2}}\log(\cos(x))\,dx$$

The second integral is a well known result easily provable with symmetry, and so we have.

$$I=-\pi\log(2)+\int_{0}^{\frac{\pi}{2}}\log(1+5\tan^2(x))\,dx$$

Now let $$\tan(x)\longrightarrow{x}$$

This gives

$$I=-\pi\log(2)+\int_{0}^{\infty}\frac{\log(1+5x^2)}{1+x^2}\,dx$$

Now let $$I(t)=\int_{0}^{\infty}\frac{\log(1+tx^2)}{1+x^2}\,dx$$

Differentiating with respect to t yields

$$I’(t)=\int_{0}^{\infty}\frac{x^2}{(1+x^2)(1+tx^2)}\,dx$$

$$=\frac{1}{t-1}[\int_{0}^{\infty}\frac{1}{1+x^2}\,dx-\int_{0}^{\infty}\frac{1}{1+tx^2}]\,dx$$

So we get

$$I’(t)=\frac{\pi}{2}\frac{1}{t-1}(1-\frac{1}{\sqrt{t}})$$

Since we know that

$$I=I(5)-I(0)-\pi\log(2) =-\pi\log(2)+\int_{0}^{5}I’(t)\,dt$$

$$I=-\pi\log(2)+\frac{\pi}{2}\int_{0}^{5}\frac{1}{t-1}(1-\frac{1}{\sqrt{t}})\,dt$$

Let $$t=x^2$$ and so

$$I=-\pi\log(2)+\pi\int_{0}^{\sqrt{5}}\frac{1}{x+1}\,dx$$

$$=\pi\log(\phi)$$

And so we achieve the desired result, namely:

$$\int_{0}^{\frac{\pi}{2}}\log(1+4\sin^2(x))\,dx=\pi\log(\phi)$$

Answer 1

Alternatively \begin{align} &\int_{0}^{\frac{\pi}{2}}\ln(1+4\sin^2x)\,dx\\ =&\int_{0}^{\frac\pi2}\int_0^4\frac{\sin^2x}{1+y\sin^2x}dy\,dx= \int_0^4\int_{0}^{\frac\pi2} \frac1y\left(1-\frac{\csc^2x}{1+y+\cot^2x}\right)dx dy\\ =&\ \frac\pi2\int_0^4 \frac1y \bigg(1-\frac1{\sqrt{1+y}}\bigg) dy =\pi\ln\left(1+\sqrt{1+y}\right)\bigg|_0^4=\pi\ln\phi \end{align}

Answer 2

By Feynman’s trick, we let $$I(a)=\int_0^{\frac{\pi}{2}} \ln \left(1+a \sin ^2 x\right) d x$$ where $a>-1$ and $I(0)=0$.

Differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+a \sin ^2 x} d x \\ & =\frac{1}{a} \int_0^{\frac{\pi}{2}} \frac{\left(1+a \sin ^2 x\right)-1}{1+a \sin ^2 x} d x \\ & =\frac{1}{a}\left[\frac{\pi}{2}-\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x+a \tan ^2 x} d x\right]\\&= \frac{1}{a}\left(\frac{\pi}{2}-\int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{1+(1+a) \tan ^2 x}\right)\\&= \frac{1}{a}\left(\frac{\pi}{2}-\frac{1}{\sqrt{1+a}}\left[\tan ^{-1}(\sqrt{1+a} \tan x)\right]_0^{\frac{\pi}{2}}\right)\\&= \frac{\pi}{2}\left(\frac{1}{a}-\frac{1}{a \sqrt{1+a}}\right) \end{aligned} $$ Integrating $I’(a)$ from $a=0$ to $4$ yields $$ I(4)-I(0)=\frac{\pi}{2} \int_0^4\left(\frac{1}{t}-\frac{1}{t{\sqrt{1+t}}}\right) d t $$ We can now conclude that $$ \boxed{I=I(4)=\pi[\ln (\sqrt{a+1}+1)]_0^4=\pi \ln \phi \, } $$

Answer 3

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} \ln \left(1+4 \sin ^2(x)\right) d x = & \int_0^{\frac{\pi}{2}} \ln (3-2 \cos 2 x) d x \\ \stackrel{2x\mapsto x}{=} & \frac{1}{2} \int_0^\pi \ln (3-2 \cos x) d x \end{aligned} $$ By my post, we get $$ \boxed{I =\frac{\pi}{2} \ln \left(\frac{3+\sqrt{9-4}}{2}\right)=\pi \ln \phi \,} $$

Answer 4

By Contour Integration

By OP, we have

$$ I=-\pi \ln 2+\int_0^{\infty} \frac{\ln \left(1+5 x^2\right)}{1+x^2} d x $$

By symmetry, we convert the integral into $$ \int_0^{\infty} \frac{\ln \left(1+5 x^2\right)}{1+x^2} d x=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+5 x^2\right)}{1+x^2} d x $$

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 letting $R \rightarrow \infty$, we have $$ \int_{-\infty}^{\infty} \frac{\ln \left(1+5 x^2\right)}{1+x^2} d x = \frac{1}{2} \oint_\gamma \frac{\ln \left(1+5 z^2\right)}{1+z^2} d z = \Re \oint_\gamma \frac{\ln (1-\sqrt{5} i z)}{1+z^2} d z $$ the last step uses the property: $\ln \left(a^2+b^2\right)=2 \Re(\ln (a+b i)).$ $$\oint_\gamma \frac{\ln (1-\sqrt{5} i z)}{1+z^2} d z =2 \pi i \lim _{z \rightarrow i} \frac{\ln (1-\sqrt{5} i z)}{2 z} = 2 \pi i \frac{\ln (1+\sqrt{5})}{2 i} = \pi \ln (1+\sqrt{5})$$ Combining yields $$ \boxed{I=\pi \ln \left(\frac{1+\sqrt{5}}{2}\right)=\pi \ln \phi}\, $$

Answer 5

$$ \begin{align} I &:= \int_{0}^{\frac{\pi}{2}}\ln\left(1+4\sin^{2}x\right)dx \\ &= \frac{1}{2}\int_{0}^{\pi}\ln\left(1+4\sin^{2}x\right)dx \\ &= \frac{1}{2}\int_{0}^{\pi}\int_{0}^{1}\frac{4\sin^{2}x}{1+4t\sin^{2}x}dtdx \\ &= 2\int_{0}^{1}\int_{0}^{\pi}\frac{\sin^{2}x}{1+4t\sin^{2}x}dxdt \\ &= \frac{1}{2}\int_{0}^{1}\int_{0}^{\pi}\frac{\left(e^{2ix}-1\right)^{2}}{te^{4ix}-\left(2t+1\right)e^{2ix}+t}dxdt \\ &= \frac{1}{4i}\int_{0}^{1}\oint_C\frac{\left(z-1\right)^{2}}{z\left(tz^{2}-\left(2t+1\right)z+t\right)}dzdt \\ &= \frac{2\pi i}{4i}\int_{0}^{1}\left(\operatorname{Res}\left(\frac{\left(z-1\right)^{2}}{z\left(tz^{2}-\left(2t+1\right)z+t\right)}, z = 0\right)+\operatorname{Res}\left(\frac{\left(z-1\right)^{2}}{z\left(tz^{2}-\left(2t+1\right)z+t\right)}, z = z_1\right)\right)dt \\ &= \frac{\pi}{2}\int_{0}^{1}\left(\lim_{z \to 0}\left(\frac{(z-0)\left(z-1\right)^{2}}{z\left(tz^{2}-\left(2t+1\right)z+t\right)}\right)+\lim_{z \to z_1}\left(\frac{(z-z_1)\left(z-1\right)^{2}}{z\left(tz^{2}-\left(2t+1\right)z+t\right)}\right)\right)dt \\ &= \frac{\pi}{2}\int_{0}^{1}\left(\frac{1}{t}-\frac{1}{t\sqrt{4t+1}}\right)dt \\ &= \frac{\pi}{2}\int_{0}^{1}\left(\frac{\sqrt{4t+1}-1}{t\sqrt{4t+1}}\right)dt \\ &= \pi\int_{1}^{\sqrt{5}}\frac{1}{u+1}du \\ &= \pi\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{align} $$

Short explanations: $C$ is the unit circle traversed in the positive direction; the simple poles on the disc of $C$ are $z = 0$ and $\displaystyle z_1 = 1+\frac{1}{2t}-\frac{\sqrt{1+4t}}{2t}$; the simple pole outside $C$ is $\displaystyle z_2 = 1+\frac{1}{2t}+\frac{\sqrt{1+4t}}{2t}$.

Answer 6

As @C-RAM suggested, to avoid convergence issues make the transformation, $$\begin{align*}\int_{0}^{\pi/2}\log(1+4\sin^2(x))\ dx&=\int_{0}^{\pi/2}\log(5)+\log\left(1-\frac{4}{5}\cos^2(x)\right)\ dx \\&=\frac{\pi\log(5)}{2}+\int_0^{\pi/2}\log\left(1-\frac{4}{5}\cos^2(x)\right)\ dx.\end{align*}$$ Expanding the logarithm by it's power series and using properties of the Beta function $B(m,n)$, $$\begin{align*}\int_0^{\pi/2}\log\left(1-\frac{4}{5}\cos^2(x)\right)\ dx&=\int_0^{\pi/2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\frac{(-4)^k}{5^k}\cos^{2k}(x)\ dx\ \\&=-\sum_{k=1}^\infty\frac{4^k}{5^kk}\int_0^{\pi/2}\cos^{2k}(x)\ dx\ \\&=-\frac{1}{2}\sum_{k=1}^\infty\frac{4^k}{5^kk}B\left(\frac{1}{2},k+\frac{1}{2}\right)\ \\&=-\frac{\sqrt\pi}{2}\sum_{k=1}^\infty\frac{4^k}{5^kk}\frac{\Gamma\left(k+1/2\right)}{\Gamma(k+1)} \ \\&=-\frac{\pi}{2}\sum_{k=1}^\infty\binom{2k}{k}\frac{1}{5^kk} \end{align*}$$ here I used, $$\int_0^{\pi/2}\cos^{2k}(x)\ dx=\frac{1}{2}B\left(\frac{1}{2},k+\frac{1}{2}\right),\quad\frac{\Gamma(k+1/2)}{\Gamma(k+1)}=\frac{\sqrt\pi}{4^k}\binom{2k}{k},$$ the first is by relation $(14)$ and the second is a well known result, also derived here. Then by the power series, $$\sum_{k=1}^\infty\binom{2k}{k}\frac{x^k}{4^kk}=2\log\left(\frac{2}{1+\sqrt{1-x}}\right)$$ taking $x=4/5$ yields, $$\sum_{k=1}^\infty\binom{2k}{k}\frac{1}{5^kk}=2\log\left(\frac{2}{1+\sqrt{1/5}}\right)$$ hence, $$\begin{align*}\int_{0}^{\pi/2}\log(1+4\sin^2(x))\ dx&=\frac{\pi\log(5)}{2}+\int_0^{\pi/2}\log\left(1-\frac{4}{5}\cos^2(x)\right)\ dx \\ &=\frac{\pi\log(5)}{2}-\pi\log\left(\frac{2}{1+\sqrt{1/5}}\right) \\ &=\pi\log\left(\frac{1+\sqrt5}{2}\right) \\ &=\pi\log(\phi).\end{align*}$$