오늘의 계산 57 - Calculation Suite

그동안 여기저기 싸지른 잡다한 계산들을 한 번 정리해볼까 합니다.

 

Problem #. Show that

\begin{equation*} \int_{0}^{\frac{\pi}{2}} \log \left(x^{2} + \log^{2}(\cos x) \right) \, \mathrm{d}x = \pi \log \log 2. \tag{1} \end{equation*}

This problem is from [IS1].

 

Solution. Let $I$ denote the integral in $\text{(1)}$. By recalling the identity

\begin{align*} x^{2} + \log^{2} (\cos x) = \left| \log \left( \frac{1+e^{2ix}}{2} \right) \right|^{2}, \end{align*}

we have

\begin{align*} I &= 2 \Re \int_{0}^{\frac{\pi}{2}} \log \log \left( \frac{1 + e^{2ix}}{2} \right) \, \mathrm{d}x = 2 \Im \int_{C} \frac{\log \log z}{2z - 1} \, \mathrm{d}z, \end{align*}

where $ z = \frac{1 + e^{2ix}}{2} $ and the contour $C$ denotes an upper semicircular contour joining from $1$ to $0$.

 

Now we know that, choosing the standard branch cut of the logarithm, the function $\log \log z$ is analytic on the upper half-plane (Indeed, $\log \log z$ is analytic on $ \Bbb{C} \setminus (-\infty, 1] $) and $\log \log (t + 0^{+}i) = \log \left| \log t \right| + i\pi$ for $0 < t < 1$. Thus by tweaking the path $C$ to

we find that

\begin{align*} I &= 2 \Im \lim_{\epsilon \to 0} \left( \int_{L_{1} + L_{2}} \frac{\log \log z}{2z - 1} \, \mathrm{d}z + \int_{\gamma_{\epsilon}} \frac{\log \log z}{2z - 1} \, \mathrm{d}z \right) \\ &= 2 \Im \left( \mathrm{PV} \int_{0}^{1} \frac{\log \left| \log t \right| + i\pi}{2t - 1} \, \mathrm{d}t + \frac{i \pi}{2} \left( \log \left| \log \frac{1}{2} \right| + i \pi \right) \right) \\ &= 2 \Im \left( \mathrm{PV} \int_{0}^{1} \frac{\log \left| \log t \right|}{2t - 1} \, \mathrm{d}t + \frac{i \pi}{2} \log \log 2 - \frac{\pi^{2}}{2} \right) \\ &= \pi \log \log 2. \end{align*}

as desired.

 

Problem #. Show that

\begin{equation*} \int_{-\infty}^{\infty} \frac{\log(1+e^{ax})}{1+e^{bx}}\, \mathrm{d}x = \frac{\pi^{2}}{12} \frac{a^{2} + b^{2}}{a b^{2}}. \tag{2} \end{equation*}

This problem is from [IS2].

 

Solution. Let $I$ denote the integral in $\text{(2)}$. With the substitution $ t = e^{bx} $, we have

\begin{align*} I &= \int_{-\infty}^{\infty} \frac{\log(1+e^{ax})}{1+e^{bx}}\, \mathrm{d}x = \frac{1}{b} \int_{0}^{\infty} \frac{\log(1+t^{a/b})}{t(1+t)}\, \mathrm{d}t. \end{align*}

Then with the substitution $ t \mapsto \frac{1}{t} $, it follows that

\begin{align*} I &= \frac{1}{b} \int_{0}^{\infty} \frac{\log(1+t^{a/b}) - \frac{a}{b} \log t}{1+t}\, \mathrm{d}t. \end{align*}

Summing two identities above, we have

\begin{align*} 2I &= \frac{1}{b} \int_{0}^{\infty} \left( \frac{\log(1 + t^{a/b})}{t} - \frac{a}{b} \frac{\log t}{1 + t} \right) \, \mathrm{d}t \\ &= \lim_{R\to\infty} \frac{1}{b} \int_{0}^{R} \left( \frac{\log(1 + t^{a/b})}{t} - \frac{a}{b} \frac{\log t}{1 + t} \right) \, \mathrm{d}t. \end{align*}

Now we focus on the integral inside the limit. Simplifying, we have

\begin{align*} & \frac{1}{b} \int_{0}^{R} \left( \frac{\log(1 + t^{a/b})}{t} - \frac{a}{b} \frac{\log t}{1 + t} \right) \, \mathrm{d}t \\ &= \frac{1}{a} \int_{0}^{R^{a/b}} \frac{\log(1 + t)}{t} \, \mathrm{d}t - \frac{a}{b^{2}} \int_{0}^{R} \frac{\log t}{1 + t} \, \mathrm{d}t \\ &= \frac{1}{a} \int_{0}^{R^{a/b}} \frac{\log(1 + t)}{t} \, \mathrm{d}t - \frac{a}{b^{2}} \log R \log (R+1) + \frac{a}{b^{2}} \int_{0}^{R} \frac{\log (1+t)}{t} \, \mathrm{d}t \\ &= -\frac{1}{a} \mathrm{Li}_{2}(-R^{a/b}) - \frac{a}{b^{2}} \mathrm{Li}_{2} (-R) - \frac{a}{b^{2}} \log R \log (R+1). \end{align*}

But we know that the dilogarithm satisfies

\begin{align*} -\mathrm{Li}_{2}(-x) &= \mathrm{Li}_{2}\left( \frac{x}{1+x} \right) + \frac{1}{2}\log^{2}(x + 1) = \zeta(2) + \frac{1}{2}\log^2 x + o(1) \end{align*}

as $ x \to \infty $. Plugging this identity above, it follows that

\begin{align*} \frac{1}{b} \int_{0}^{R} \left( \frac{\log(1 + t^{a/b})}{t} - \frac{a}{b} \frac{\log t}{1 + t} \right) \, \mathrm{d}t &= \frac{a^2 + b^2}{a b^2} \zeta(2) + o(1) \end{align*}

as $ R \to \infty $, and therefore the proof is complete.

References

1. Shobhit, Problem 23 of Integration Contest - Season 3 in "Integrals and Series"
2. zaidalyafey, a mysterious integral in "Integrals and Series"