Evaluate the definite integral $\int^{\pi/2}_0 \ln(1+p \cos(x)) dx $

Question

I need help to evaluate :$$I(p)=\int^{\pi/2}_0 \ln(1+p \cos(x)) dx $$ therfore: $$I'(p)=\int^{\pi/2}_0 \frac{\cos(x)}{1+p\cos(x)}dx=\frac{1}{p}\int^{\pi/2}_0\left(1-\frac{1}{1+p\cos(x)}\right)dx$$

$$I'(p)\overset{y=\tan(x/2)}{=}\frac{\pi}{2p}-\frac{2}{p}\int^{1}_0 \frac{dy}{(1-p)y^2+(1+p)}$$

therfore: $$I'(p)=\frac{\pi}{2p}-\frac{2}{p}\sqrt{\frac{1-p}{1+p}}\arctan\left({\sqrt{\frac{1-p}{1+p}}}\right)$$

Therfore: $$I(b)=\frac{\pi}{2}\ln(p)-\int^p_0 \frac{2}{y}\sqrt{\frac{1-y}{1+y}}\arctan\left({\sqrt{\frac{1-y}{1+y}}}\right)dy$$

Can this last integral be calculated? Is there another way to evaluate this integral?

Answer 1

Let $p=\sin \theta$, with $ \int_0^\theta\frac{y}{\sin y}dy =2\text{Ti}_2(\tan\frac\theta 2 )$, to arrive at the close-form below \begin{align} &\int_0^\frac{\pi}{2} \ln(1+p \sin x)\ dx\\ =& \int_0^\frac{\pi}{2} \ln(1+\sin \theta\sin x)\ dx =\int_0^\frac{\pi}{2}\int_0^\theta\frac{\cos y \sin x}{1+\sin y\sin x}dy\ dx\\ =& \int_0^\theta \cot y\int_0^\frac{\pi}{2} \bigg(1- \frac{1}{1+\sin y\sin x}\bigg)dx\ dy = \int_0^\theta \cot y \bigg( \frac\pi2 -\frac{\frac\pi2-y}{\cos y}\bigg)dy\\ =& \int_0^\theta \left(\frac{y}{\sin y}-\frac\pi2\tan\frac y2 \right) dy = 2 \text{Ti}_2\left(\tan\frac\theta 2 \right)+ \pi \ln\cos\frac {\theta}2 \\ =&\ 2\text{Ti}_2\bigg(\frac{1-\sqrt{1-p^2}}p\bigg)+ \frac\pi2\ln\frac{p+\sqrt{1-p^2}}{2} \end{align}

Answer 2

The part of $I'(p)$ difficult to integrate is $$\frac{2}{p \sqrt{1 - p^2}} \arctan \sqrt \frac{1 - p}{1+p}$$

Substituting $p = \cos \theta$ transforms the integral of that quantity w.r.t. $p$ to $$\int \theta \sec \theta \,d\theta ,$$ which has no antiderivative expressible in closed form in terms of elementary functions, but which can be expressed in terms of the dilogarithm function, $\operatorname{Li}_2$, or the inverse tangent integral function, $\operatorname{Ti}_2$, e.g., as $$\theta \log \frac{1 - i e^{i \theta}}{1 + i e^{i \theta}} + i (\operatorname{Li}_2(1 + i e^{i \theta}) - \operatorname{Li}_2(1 - i e^{i \theta})) + C ;$$ the first term in the expression can alternatively be expressed in terms of $\arctan \exp i \theta$.

For certain values of $p$ the original integral simplifies nicely, e.g., $$I(\pm 1) = - \frac\pi2 \log 2 \pm 2 G,$$ where $G$ is Catalan's constant.

See this question for details and alternative expressions for the antiderivative.

Remark (asymptotic formulae)

• Using that $I(-1) = -\frac\pi2 \log 2 - 2 G$ and expanding the formula for $I'(p)$ in a series gives an asymptotic expansion for $I(p)$ as $p \searrow -1$: $$I(p) = -\frac\pi2 \log 2 - 2 G + \sqrt 2 \pi \sqrt{p - (-1)} - \left(1 + \frac\pi2\right) (p - (-1)) + \cdots,$$ where $\cdots$ denotes a remainder in $O\left((p - (-1))^{\frac32}\right)$.
• Using that $I(0) = 0$ and expanding the formula for $I'(p)$ in a Taylor series gives an asymptotic expansion as $p \to 0$: $$I(p) = p - \frac\pi8 p^2 + \frac29 p^3 + \cdots,$$ where $\cdots$ now denotes a remainder in $O(p^4)$.
• For large $p$, the $1$ in the argument of $\log$ makes proportionally little difference, so $$I(p) \approx \int_0^{\frac \pi2} \log (p \cos x) \,dx = \int_0^{\frac \pi2} \log p \,dx + \int_0^{\frac \pi2} \log \cos x \,dx = \frac\pi2 (\log p - \log 2) $$ (The last equality uses a well-known identity.) Analyzing the difference gives that $I(p)$ has asymptotic expansion $$I(p) = \frac\pi2 \log p - \frac\pi2 \log 2 + \frac{\log p}p + (1 + \log 2) \frac1p + \cdots ,$$ where $\cdots$ now denotes a remainder in $O\left(\frac1{p^2}\right)$.

Answer 3

Using the tangent half angle substitution $$I=\int^{\pi/2}_0 \log(1+p \cos(x))\, dx=2\int_0^1 \frac 1{1+t^2} \log\Big(1+p \,\frac{1-t^2}{1+t^2} \Big)$$ Write the logarithm as $$\log \left(t-\frac{\sqrt{p+1}}{\sqrt{p-1}}\right)+\log \left(t+\frac{\sqrt{p+1}}{\sqrt{p-1}}\right)+\log (1-p)-\log \left(1+t^2\right)$$ and eventually $$\frac 1{1+t^2}=\frac i 2\left(\frac{1}{t+i}-\frac{1}{t-i}\right)$$ to face simple integrals.

Recombining, this leads to

$$I=2 C+\frac{1}{2} \pi \log \left(\frac{p}{2}\right)+i\,(A-B)$$ where $$A=\text{Li}_2\left(\frac{(1+i) \sqrt{p-1}}{\sqrt{p-1}-i \sqrt{p+1}}\right)+\text{Li}_2\left(\frac{(1+i) \sqrt{p-1}}{\sqrt{p-1}+i \sqrt{p+1}}\right)$$ $$B=\text{Li}_2\left(\frac{(1+i) \sqrt{p-1}}{i \sqrt{p-1}-\sqrt{p+1}}\right)+\text{Li}_2\left(\frac{(1+i) \sqrt{p-1}}{i \sqrt{p-1}+\sqrt{p+1}}\right)$$

Expanded for large values of $p$, thies gives $$I=\frac{1}{2} \pi \log \left(\frac{p}{2}\right)+\sum_{n=0}^\infty \frac {a_n}{p^{2n+1}}\tag 1$$ Using $L=\log(2p)$ , the first coefficients are $$\large\left( \begin{array}{cc} n & a_n \\ 0 & L+1 \\ 1 & \frac{6 L-1}{36} \\ 2 & \frac{60 L-23}{800} \\ 3 & \frac{420 L-199}{9408} \\ 4 & \frac{2520 L-1319}{82944} \\ 5 & \frac{27720 L-15377}{1239040} \\ \end{array} \right)$$ which is a good approximation. For example, for $p=2$, the above truncated series leads to $$\frac{24948591219443}{50357757542400}+\frac{951195037 }{908328960}\log (2)=\color{red}{1.22128}53$$

Edit

May be interesting if to notice that, using $$I(p)=\int^{\pi/2}_0 \log(1+p \cos(x))\, dx$$ $$I'(p)=\int^{\pi/2}_0 \frac{\cos (x)}{1+p \cos (x)}\,dx=\frac{\pi }{2 p}-\frac{2}{p \sqrt{p^2-1}}\tanh^{-1}\left(\sqrt{\frac{p-1}{p+1}}\right)$$

Expanding for large values of $p$ $$I'(p)=\frac{\pi }{2 p}-\sum_{n=1}^\infty \frac {b_n}{p^{2n}}\tag 2$$ the first coefficients being (still with $L=\log(2p)$) $$\large\left( \begin{array}{cc} n & b_n \\ 1 & L \\ 2 & \frac{2 L-1}{4} \\ 3 & \frac{12 L-7}{32} \\ 4 & \frac{60 L-37}{192} \\ 5 & \frac{840 L-533}{3072} \\ \end{array} \right)$$

Integrating from $p=1$ to $p$, the results of $(1)$.