I want to integrate the following integral using a variety of methods. It came up while I was working out a solution for $\int_0^\pi\frac{x(\pi-x)}{\sin{(x)}}$. $$\int_0^\infty\frac{\ln{(x)}\ln{(1+ix)}}{1+x^2}dx$$ Using the Taylor series, I have tried to expand $\ln{(1+ix)}$. I tried to use $\int_0^\infty\frac{x^n\ln{(1+ix)}}{x^2+1}dx$ to solve the above problem, but I couldn't solve this one either. I tried integration by parts by setting $u=\ln{(x)}\ln{(1+ix)}$ and $dv=\frac{1}{x^2+1}$ but it seemed to complicate it further. I'm not acquainted enough with complex analysis to solve it that way, but I would appreciate a complex analysis answer regardless. I'm not sure what else I can do. Thank you in advance
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- $\begingroup$ "... by a variety of methods." Do you really want multiple solutions, or will one do? $\endgroup$Thomas Andrews– Thomas Andrews2022-10-15 01:08:08 +00:00Commented Oct 15, 2022 at 1:08
- $\begingroup$ @ThomasAndrews I only need one $\endgroup$chroma24– chroma242022-10-15 01:09:36 +00:00Commented Oct 15, 2022 at 1:09
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3 Answers
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Note
\begin{align} &\int_0^\infty\frac{\ln{x}\ln{(1+ix)}}{1+x^2}dx\\ =&\ \frac12 \int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}dx + i\int_0^\infty\frac{\ln{x}\tan^{-1}x}{1+x^2}dx \end{align} where \begin{align} \int_0^\infty\frac{\ln{x}\ln{(1+x^2)}}{1+x^2}{dx}& \overset{x\to \frac1x}=\int_0^\infty\frac{\ln^2{x}}{1+x^2}dx =\frac{\pi^3}8\\ \\ \int_0^\infty \frac{\ln x\tan^{-1}x}{1+x^2}dx =& \int_0^\infty \int_0^1 \frac{x\ln x}{(1+x^2)(1+y^2x^2)} \overset{x\to \frac1{xy}}{dx}dy\\ = & \ \frac1{2}\int_0^1\int_0^\infty \frac{-x\ln y}{(1+x^2)(1+{y^2}x^2)} {dx}\ dy\\ =& \ \frac12\int_0^1\frac{\ln^2 y}{1-y^2}dy =\frac78\zeta(3) \end{align}
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2 Using the fact that $$ \begin{aligned} \ln (1+i x) &=\ln \left(\sqrt{1+x^2} \cdot e^{i \tan ^{-1} x}\right) \\ &=\frac{1}{2} \ln \left(1+x^2\right)+i \tan ^{-1}x, \end{aligned} $$ we splits the integral into two as $$ I=\frac{1}{2} \underbrace{\int_0^{\infty} \frac{\ln x \ln \left(1+x^2\right)}{1+x^2}}_J d x+i \underbrace{\int_0^{\infty} \frac{\ln x \tan ^{-1} x}{1+x^2} d x}_K $$ Letting $x\mapsto\tan x$ yields $$ \begin{aligned} J &=-2 \int_0^{\frac{\pi}{2}} \ln (\tan x) \ln (\cos x) d x \\ &=2 \int_0^{\frac{\pi}{2}} \ln ^2(\cos x)dx-2 \int_0^{\frac{\pi}{2}} \ln (\sin x) \ln (\cos x) d x \\ &=2 \cdot \frac{1}{24}\left(\pi^3+3 \pi \ln ^2 4\right)-2\left(-\frac{\pi^3}{48}+\frac{\pi}{2} \ln ^2 2\right)\cdots (*) \\ &=\frac{\pi^3}{8} \end{aligned} $$ where $(*)$ comes from my post .
$$ \begin{aligned} K &=\int_0^{\infty} \frac{\ln x \tan ^{-1} x}{1+x^2} d x=\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\frac{7}{8} \zeta(3), \end{aligned} $$ where the last result comes from my post.
We can now conclude that $$ \boxed{I=\frac{1}{16}\left(\pi^3+14 i\zeta(3)\right)} $$
- $\begingroup$ I think you forget to take the $1/2$ in front of the $J$ integral into account. $\endgroup$John Barber– John Barber2022-10-15 17:09:09 +00:00Commented Oct 15, 2022 at 17:09
- $\begingroup$ Thank you very much for catching it. Fixed. $\endgroup$Lai– Lai2022-10-15 22:14:35 +00:00Commented Oct 15, 2022 at 22:14
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3 Let $$I = \int_0^\infty\frac{\ln(x)\ln(1+ix)}{1+x^2}\,dx$$
By substituting $x\mapsto\frac1x$, we have
$$\int_1^\infty \frac{\ln(x)\ln(1+ix)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)\ln\left(1+\frac ix\right)}{1+\frac1{x^2}} \, \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)(\ln(x+i)-\ln(x))}{1+x^2} \, dx$$
Rejoining this with the part of $I$ over $[0,1]$, we get
$$\begin{align*} I &= \int_0^1 \frac{\ln(x)}{1+x^2} \ln\left(\frac{1+ix}{i+x}\right) \, dx + \int_0^1 \frac{\ln^2(x)}{1+x^2} \, dx \\[1ex] &= \int_0^1 \frac{\ln(x)}{1+x^2}\ln\left(\frac{1+ix}{1-ix}\right) \, dx - \ln(i) \int_0^1 \frac{\ln(x)}{1+x^2} \, dx + \int_0^1 \frac{\ln^2(x)}{1+x^2} \, dx \\[1ex] &= 2i \int_0^1 \frac{\ln(x)\arctan(x)}{1+x^2} \, dx - \frac{i\pi}2 \int_0^1 \frac{\ln(x)}{1+x^2} \, dx + \int_0^1 \frac{\ln^2(x)}{1+x^2} \, dx \end{align*}$$
For the latter two integrals, we consider
$$J_a = \int_0^1 \frac{\ln^a(x)}{1+x^2} \, dx \\ K_a = \int_0^1 x^{2n} \ln^a(x) \, dx$$
Derive some recurrences:
$$\begin{align*} J_a &= -a \int_0^1 \frac{\ln^{a-1}(x) \arctan(x)}x \, dx \tag{1} \\[1ex] &= -a \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{2n} \ln^{a-1}(x) \, dx \tag{2} \\[1ex] &= -a \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} K_{a-1} \\[3ex] K_{a-1} &= -(a-1) \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \int_0^1 x^{2n} \ln^{a-2}(x) \, dx \tag{1} \\[1ex] &= -(a-1) \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} K_{a-2} \\[3ex] \implies J_a &= (-1)^a a! \beta(a+1) \tag{3} \\[3ex] \implies I &= 2i \int_0^1 \frac{\ln(x)\arctan(x)}{1+x^2} \, dx + \frac{i\pi}2 G + \frac{\pi^3}{16} \end{align*}$$
For the remaining integral,
$$\begin{align*} \int_0^1 \frac{\ln(x)\arctan(x)}{1+x^2} \, dx &= \int_0^{\frac\pi4} x \ln(\tan(x)) \, dx \tag{4} \\[1ex] &= \int_0^{\frac\pi4} x \ln(\sin(x)) \, dx - \int_0^{\frac\pi4} x \ln(\cos(x)) \, dx \\[1ex] &= \int_{\frac\pi4}^{\frac\pi2} \left(\frac\pi2-x\right) \ln\left(\sin\left(\frac\pi2-x\right)\right) \, dx - \int_0^{\frac\pi4} x \ln(\cos(x)) \, dx \tag{5} \\[1ex] &= \frac\pi2 \int_0^{\frac\pi4} \ln(\sin(x)) \, dx - \int_0^{\frac\pi2} x \ln(\cos(x)) \, dx \\[1ex] &= -\frac\pi4 G - \frac{\pi^2}8 \ln(2) - \int_0^{\frac\pi2} x \ln(\cos(x)) \, dx \tag{6} \end{align*}$$
I did this last integral with Fourier series, but you might find more streamlined methods here.
$$\begin{align*} \int_0^{\frac\pi2} x \ln(\cos(x)) &= \int_0^{\frac\pi2} \left(\frac\pi2 - x\right) \ln\left(\cos\left(\frac\pi2-x\right)\right) \, dx \tag{5} \\[1ex] &= \frac\pi2 \int_0^{\frac\pi2} \ln(\sin(x)) \, dx - \int_0^{\frac\pi2} x \ln(\sin(x)) \, dx \\[1ex] &= \frac\pi2 \int_0^{\frac\pi2} \ln\left(\sin\left(\frac\pi2-x\right)\right) \, dx - \int_0^{\frac\pi2} x \ln(\sin(x)) \, dx \tag{5} \\[1ex] &= \frac\pi2 \int_0^{\frac\pi2} \ln(\cos(x)) \, dx - \int_0^{\frac\pi2} x \ln(\sin(x)) \, dx \\[1ex] &= -\frac{\pi^2}4\ln(2) - \int_0^{\frac\pi2} x \ln(\sin(x)) \, dx \tag{7} \\[1ex] &= -\frac{\pi^2}4\ln(2) + \int_0^{\frac\pi2} x \left(\ln(2) + \sum_{k=1}^\infty \frac{\cos(2kx)}k\right) \, dx \\[1ex] &= -\frac{\pi^2}4 \ln(2) + \ln(2) \int_0^{\frac\pi2} x \, dx + \sum_{k=1}^\infty \frac1k \int_0^{\frac\pi2} x \cos(2kx) \, dx \\[1ex] &= -\frac{\pi^2}8\ln(2) - \sum_{k=1}^\infty \frac1{2k^2} \int_0^{\frac\pi2} \sin(2kx) \, dx \tag{1} \\[1ex] &= -\frac{\pi^2}8\ln(2) + \frac14 \sum_{k=1}^\infty \frac{(-1)^k - 1}{k^3} \\[1ex] &= -\frac{\pi^2}8\ln(2)-\frac7{16}\zeta(3) \end{align*}$$
Putting everything together, we conclude that
$$\begin{align*} I &= 2i\left(-\frac\pi4 G - \frac{\pi^2}8 \ln(2) - \left(-\frac{\pi^2}8\ln(2)-\frac7{16}\zeta(3)\right)\right) + \frac{i\pi}2 G + \frac{\pi^3}{16} \\[1ex] &= \boxed{\frac{\pi^3+14\zeta(3)\,i}{16}} \end{align*}$$
- $(1)$ : integrate by parts
- $(2)$ : exploit the series expansion of $\arctan(x)$
- $(3)$ : see Dirichlet beta
- $(4)$ : substitute $x\mapsto\tan(x)$
- $(5)$ : substitute $x\mapsto\frac\pi2-x$
- $(6)$ : see the integral $I$ in this answer
- $(7)$ : see here
- $\begingroup$ We may also be able to integrate $\frac{\ln(x)\arctan(x)}{1+x^2}$ by noting$$\int_0^1\frac{\ln(x)\arctan(x)}{1+x^2}\,dx=\int_0^\infty\frac{\ln(x)\arctan(x)}{1+x^2}\,dx+\frac\pi2\int_0^1\frac{\ln(x)}{1+x^2}\,dx$$and use the same contour demonstrated [here](math.stackexchange.com/a/4561985/170231) to attack the first integral on the right side. This time, we'd take$$I(a)=\frac1{2i}\int_0^\infty\frac{\log(x)\log\left(\frac{i-x}{i+x}\right)}{a^2+x^2}\,dx$$though I haven't checked if this is fruitful. $\endgroup$user170231– user1702312022-11-23 00:45:18 +00:00Commented Nov 23, 2022 at 0:45
- 1$\begingroup$ creative answer! I would have solved $J_a$ by taking $a$ derivatives of $\int_0^1 \frac{x^n}{1+x^2} dx$ with respect to n and evaluate at n=0 $\endgroup$chroma24– chroma242022-11-23 02:15:54 +00:00Commented Nov 23, 2022 at 2:15
- $\begingroup$ @phi-rate You can find a complex-analysis solution here for a slightly different version of the integral $I(a)$ proposed here. $\endgroup$user170231– user1702312022-11-27 23:31:35 +00:00Commented Nov 27, 2022 at 23:31