# Integral $\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}$

I have stumbled upon the following integral:$$I=\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^2} \frac{dx}{\sqrt x}=-\frac{\pi}{24}$$ Although I could solve it, I am not quite comfortable with the way I did it.

But first I will show the way. We can substitute $$\ln x \rightarrow t\$$ which gives: $$I=\int_{-\infty}^\infty \frac{t}{\pi^2+t^2}\frac{e^{\frac{t}{2}}}{(1+e^t)^2}dt\overset{t=-x}=\int_{-\infty}^\infty \frac{-x}{\pi^2+x^2}\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}dx$$ Also adding the two integral from above and simplify some of it yields: $$2I= \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\left(\frac{e^{\frac{x}{2}}}{(1+e^x)^2}-\frac{e^{-\frac{x}{2}}}{(1+e^{-x})^2}\right)dx$$ $$\Rightarrow I=-\frac{1}{4} \int_{-\infty}^\infty \frac{x}{\pi^2+x^2}\frac{\sinh \left(\frac{x}{2}\right)}{\cosh ^2\left(\frac{x}{2}\right)}dx$$ And now a round of IBP gives: $$I=\frac12 \int_{-\infty}^\infty \left(\frac{x^2-\pi^2}{(x^2+\pi^2)^2}\right)\left(\frac{1}{\cosh \left(\frac{x}{2}\right)}\right)dx$$ Using the Plancherel theorem the integral simplifies to: $$I=\int_0^\infty \left(\sqrt{\frac{\pi}{2}}x\left(-e^{-\pi x}\right)\right)\left(\sqrt{2\pi}\frac{1}{\cosh(\pi x)}\right)dx\overset{\pi x\rightarrow x}=-\frac{1}{\pi}\int_0^\infty \frac{x}{\cosh( x)}e^{- x}dx$$ We also have the following Laplace tranform for:$$f(t)=\frac{t}{\cosh( t)}\rightarrow F(s)=\frac18\left(\psi_1\left(\frac{s+1}{4}\right)-\psi_1\left(\frac{s+3}{4}\right)\right)$$ Where $$\displaystyle{\psi_1(z)=\sum_{n=0}^\infty \frac{1}{(z+n)^2}}\,$$ is the trigamma function. $$\Rightarrow I=-\frac{1}{\pi}F(s=1)=-\frac{1}{\pi}\cdot \frac18\left(\psi_1\left(\frac{1}{2}\right)-\psi_1 (1)\right)=-\frac{1}{\pi}\cdot \frac18\left(\frac{\pi^2}{2}-\frac{\pi^2}{6}\right)=-\frac{\pi}{24}$$ Have I done anything wrong, or can it be improved? I have to admit that I mostly used wolfram when applying Plancherel theorem and Laplace transform which I'm not comfortable with, but I didn't find an alternative method myself.

For this question I would like to see a different proof that doesn't rely on that theorem.

Probably not needed, but I should mention that my contour integration knowledge is pretty low. Also maybe there is a conexion with this integral, but I didn't find any.

From the identity

$$\Im\int_0^\infty e^{-(\pi-it)x}\,dx=\frac t{\pi^2+t^2}$$

we see that it suffices to compute the imaginary part of the integral

$$\int_0^\infty dx\int_{-\infty}^\infty dt\; \frac{e^{\alpha t}}{(1+e^t)^2}e^{-\pi x}$$

where $$\alpha=1/2+ix$$. Now, the integral with respect to $$t$$ is easy by means of the substitution $$u=e^t$$ and using the beta function. We thus get

$$\int_0^\infty \pi\left(\frac12-ix\right)\frac{e^{-\pi x}}{\cosh(\pi x)}\,dx.$$

Taking the imaginary part we see that the problem boils down to compute the integral

$$\int_0^\infty \frac{x e^{-\pi x}}{\cosh(\pi x)}\,dx =2\int_0^\infty \frac{x}{1+e^{2\pi x}}\,dx$$

which, after the substitution $$v=2\pi x$$, reduces to the integral representation of the eta function $$\eta(2)$$. Also notice that taking the real part we obtain the evaluation

$$\int_0^\infty\frac1{(\pi^2+\log^2 x)(1+x)^2} \frac{dx}{\sqrt x}= \frac{\log2}{2\pi}.$$

This method generalises to other integrals such as

\begin{align*} \int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x} &=\frac{3\log (2)}{8 \pi }-\frac{3 \zeta (3)}{16 \pi ^3}\\ \int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^3} \frac{dx}{\sqrt x} &=-\frac{\pi }{24}\\ \int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x} &=\frac{5 \log (2)}{16 \pi }-\frac{9 \zeta (3)}{32 \pi ^3}\\ \int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^4} \frac{dx}{\sqrt x} &=-\frac{223 \pi }{5760}\\ \int_0^\infty \frac{1}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x} &=-\frac{43 \zeta (3)}{128 \pi ^3}+\frac{15 \zeta (5)}{256 \pi ^5}+\frac{35 \log (2)}{128 \pi }\\ \int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^5} \frac{dx}{\sqrt x} &=-\frac{103 \pi }{2880}\\ \end{align*}

Incidentally, since the integrals $$\int_0^\infty \frac{\ln x}{(\pi^2+\ln^2 x)(1+x)^k} \frac{dx}{\sqrt x}$$ both yield the same value for $$k=2,3$$, we also deduce

$$\int_0^\infty \frac{\sqrt x\ln x}{(\pi^2+\ln^2 x)(1+x)^3}\;dx=0.$$

This, however, should not be surprising due to the symmetry $$x\mapsto1/x$$.

• At first sight this looks quite impressive! Give me some time to try to understand it better. – Zacky Jan 1 '19 at 19:13
• Thanks! I hope you enjoy completing the steps, but tell me if you need more details. – diech Jan 1 '19 at 19:18
• This is very beautiful. But I get here: $$\int_0^\infty \pi\left(\frac12\color{red}+ix\right)\frac{e^{-\pi x}}{\cosh(\pi x)}\,dx.$$ Am I wrong? – Zacky Jan 1 '19 at 22:50
• @Zacky Are you sure? I've checked my calculations and I am still getting $-ix$. In any case, your integral is negative, so the imaginary part should also be negative, shouldn't it? – diech Jan 2 '19 at 9:34
• Well, I have $$\int_{-\infty}^\infty \frac{e^{at}}{(1+e^t)^2}dt\overset{t=\ln u}=\int_0^\infty \frac{u^{a-1}}{(1+u)^2}du=B(a,2-a)=a\Gamma(a)\Gamma(1-a)$$And since $a=\frac12\color{red}+ix$ and by the reflection formula it gives that. – Zacky Jan 2 '19 at 11:30

That $$\pi^2+\log^2(x)$$ makes me think to the integral representation for Gregory coefficients:

$$\int_{0}^{+\infty}\frac{dx}{(1+x)^n (\pi^2+\log^2 x)} = \frac{1}{n!}\left[\frac{d^n}{dx^n}\frac{z}{\log(1-z)}\right]_{z=0}=[z^n]\frac{z}{\log(1-z)} \tag{1}$$ which can be seen as a consequence of the Lagrange-Buhrmann inversion theorem. We just need to insert a factor $$\frac{\log x}{\sqrt{x}}$$ in the integrand function appearing in the LHS, so let's go back to the residue theorem.

$$\int_{0}^{+\infty}\frac{\log(x)\,dx}{\sqrt{x}(1+x)^2(\pi^2+\log^2 x)}=\int_{-\infty}^{+\infty}\frac{t e^{-t/2}}{(2\cosh\frac{t}{2})^2 (\pi^2+t^2)}\,dt$$ equals $$-\frac{1}{4}\int_{\mathbb{R}}\frac{t\sinh\frac{t}{2}}{(t^2+\pi^2)\cosh^2\frac{t}{2}}\,dt=-\int_{\mathbb{R}}\frac{t\sinh t}{(4t^2+\pi^2)\cosh^2 t}\,dt.$$ The meromorphic function $$\frac{\sinh t}{\cosh^2 t}=-\frac{d}{dt}\left(\frac{1}{\cosh t}\right)$$ only has double poles with residue zero, hence all the mass of the last integral comes from the singularity at $$\frac{\pi i}{2}$$ and from the behaviour at infinity. The residue theorem grants $$\frac{1}{\cosh x}=\sum_{n\geq 0}(-1)^n \frac{\pi(2n+1)}{\frac{\pi^2}{4}(2n+1)^2+x^2}$$ and $$\frac{\sinh x}{\cosh^2 x} = \sum_{n\geq 0}(-1)^n \frac{2\pi(2n+1)x}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2}.$$ Since $$\int_{\mathbb{R}}\frac{2\pi(2n+1)x^2}{(\frac{\pi^2}{4}(2n+1)^2+x^2)^2 (\pi^2+4x^2)}\,dx = \frac{1}{2\pi(n+1)^2}$$ our integral equals $$-\frac{1}{2\pi}\eta(2)=\color{red}{-\frac{\pi}{24}}$$ by the dominated convergence theorem, allowing to switch $$\int_{\mathbb{R}}$$ and $$\sum_{n\geq 0}$$. $$\frac{1}{24}$$ is also the coefficient of $$z^3$$ in $$\frac{z}{\log(1-z)}$$, but so far I have not found a direct way to relate the original integral to the $$n=3$$ instance of $$(1)$$.

• @Jack D'Aurizio, can I ask how you obtained those series? It seems you've used the Residue Theorem, but I'm not sure exactly how. Can the method be used to create series for other functions, similar to how we can compute Taylor series? – Zachary Jan 1 '19 at 19:42
• @Zachary: this is Herglotz trick for $\frac{1}{\cosh}$, followed by termwise differentiation. We have $\frac{1}{\cosh z} = \sum_{\xi} \frac{R_{\xi}}{z-\xi}$ where $\xi$ are the (simple) zeroes of $\cosh$ and $R_\xi$ is the residue at $z=\xi$ of $\frac{1}{\cosh}$. – Jack D'Aurizio Jan 1 '19 at 19:50
• @Zachary: the trick to get my representation is to couple the contribution provided by the poles at $(2n+1)\frac{\pi i}{2}$ for $n=N\in\mathbb{N}$ and $n=-(N+1)$.This cancels the imaginary part and leaves a series on $n\geq 0$. – Jack D'Aurizio Jan 2 '19 at 0:51
• @Zachary: that's my fault, I wrote wrong constants, now fixing. – Jack D'Aurizio Jan 2 '19 at 6:36
• Alright. Can I just say that the coupling of poles method is genius. Before doing that, I had an absolute mess of an infinite series, but thanks to your suggestion I managed to nicely simplify the sum. I really appreciate your answers here and I learn a lot from them. I especially enjoyed this one: math.stackexchange.com/questions/2529614/… :) Now, I'm always ready to use the Laplace "Integration by parts" methods. – Zachary Jan 2 '19 at 7:53

This is not a full answer, but one does not need the full Laplace transform of $$x/\cosh x$$, as the integral can be done by expanding $$1/\cosh x$$ into a geometric series

\begin{aligned} I = \int_{0}^{\infty}\frac{xe^{-x}}{\cosh x}\,\mathrm{d}x &= 2\int_{0}^{\infty}\frac{xe^{-x}}{e^{x}}\frac{\mathrm{d}x}{1+e^{-2x}} = 2\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}xe^{-(2+2n)x}\,\mathrm{d}x \\ &= 2\sum_{n=0}^{\infty}\frac{(-1)^{n}}{4(1+n)^{2}}\int_{0}^{\infty}ue^{-u}\,\mathrm{d}u = \frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(1+n)^{2}} \\ &= \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2}} = \frac{\eta(2)}{2} = \frac{\pi^{2}}{24}\end{aligned}

where $$\eta(s)$$ is the Dirichlet eta function.

To show that $$(-1)^{n-1} \int_{0}^{\infty}\frac{ \mathrm dx}{ (\pi^{2}+\ln^{2} x)(1+x)^{n}}$$ is an integral representation of the Gregory coefficients, some textbooks integrate the function $$\frac{1}{(\ln z - i\pi)(1+z)^{n}}$$ around a keyhole contour. We can do something similar here.

Let's integrate the function $$f(z)= \frac{1}{(\ln z - i \pi) (1+z)^{2}} \frac{1}{\sqrt{z}}$$ around a keyhole contour where the branch cut is along the positive real axis.

Integrating around the contour, we get $$\int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{\infty}^{0} \frac{1}{(\ln x + 2 \pi i - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{e^{2 \pi i}x}} = 2 \pi i \operatorname{Res} \left[f(z), -1 \right]$$

The left side of the above equation is

$$\int_{0}^{\infty}\frac{1}{(\ln x - i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} + \int_{0}^{\infty} \frac{1}{(\ln x + i \pi) (1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}}$$

$$= \int_{0}^{\infty} \frac{2 \ln x}{\left(\ln^{2}(x)+\pi^{2}\right)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}}$$

Since $$f(z)$$ has pole at $$z=e^{i \pi}=-1$$ of order three, calculating the residue of $$f(z)$$ at $$z=-1$$ is a bit tedious.

But at $$z=-1$$, $$\ln(z) - i\pi$$ has the Taylor series expansion $$\ln(z) - i \pi = - (z+1)-\frac{(z+1)^{2}}{2!}- \frac{(z+1)^{3}}{3!} + O \left((z+1)^{4} \right)$$

Using polynomial long division, one can then show that Laurent series expansion of $$\frac{1}{\ln z - i \pi}$$ about $$z=-1$$ is $$\frac{1}{\ln z - i \pi} = - \frac{1}{z+1} + \frac{1}{2} + \frac{z+1}{12} + O \left((z+1)^{2} \right)$$

So near $$z=-1$$, $$f(z) = \frac{1}{\sqrt{z}} \left(- \frac{1}{(z+1)^{3}} + \frac{1}{2(z+1)^{2}} + \frac{1}{12(z+1)} +O(1)\right),$$

from which we get \begin{align} \small \operatorname{Res}\left[f(z), -1 \right] &= \small -\operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{3}}, -1 \right] + \frac{1}{2} \, \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{(z+1)^{2}}, -1 \right] + \frac{1}{12} \, \operatorname{Res}\left[\frac{1}{\sqrt{z}} \frac{1}{z+1}, -1 \right] \\ &=- \frac{1}{2!} \frac{3/4}{(e^{i \pi})^{5/2}} - \frac{1}{2} \frac{1/2}{(e^{i \pi)^{3/2}}} + \frac{1}{12} \frac{1}{(e^{i \pi})^{1/2}} \\ &= - \frac{3}{8i} + \frac{1}{4i} + \frac{1}{12i} = -\frac{1}{24i} \end{align}

(Alternatively, we could have also expanded $$\frac{1}{\sqrt{z}}$$ at $$z=-1$$ to get the first few terms of the Laurent series expansion of $$f(z)$$ about $$z=-1$$.)

Therefore, $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+\ln^{2} x)(1+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \frac{2 \pi i}{2} \left(-\frac{1}{24i} \right) = -\frac{\pi}{24}$$

If $$a>0$$ and $$a \ne 1$$, then the same approach shows that $$\int_{0}^{\infty} \frac{\ln x}{(\pi^{2}+ \ln^{2}x)(a+x)^{2}} \frac{\mathrm dx}{\sqrt{x}} = \pi \left(\frac{2 + \ln a}{2a^{3/2} \ln^{2}a}- \frac{1}{(a-1)^{2}} \right)$$

Applying L'Hôpital's rule 4 times shows that the limit of the right side of the above equation as $$a \to 1$$ is $$-\frac{\pi}{24}$$.