Evaluate $\int_0^{+\infty} \big ( \frac{\ln x}{x - 1}\big)^2dx $ using residues and complex analysis

Question

I am trying to evaluate $\int_0^{\infty} \big ( \frac{\ln x}{x - 1}\big)^2dx $, where $\big ( \frac{\ln x}{x - 1} \big)^2 = f(x)$ with residues and complex analysis. Pole here is $x_0$ = 1 and it has second order. Contour is here.

We need to use $f(z)$ instead of $f(x)$ for complex integration and split start function $f(z)$ to 2 functions: $R(z)$ and $h(z)$, where $$R(z) = \frac{1}{(z-1)^2},$$ $$h(z) = \ln(z) + i\phi,0 <\phi < 2\pi.$$ - that was in my lectures notes.
So could you help me with next steps?

Answer 1

The function to be integrated (and any related function that will ultimately solve the problem) hates and wants to avoid the points $0,1$, so the pink contour would be a first guess to apply complex analysis (C.A. for short below):

Let $a$ be $a=2\pi i$. We quickly find a polynomial $P=P_a$, so that $P(X+a)-P(X)=X^2$, this is $$ P(X) = \frac 1{3a}\left(x^3-\frac 32ax^2+\frac 12a^2x\right) \ . $$ Then $$ \int_{\color{magenta}{\text{Pink contour}}} \frac {P(\ln z)}{(z-1)^2} =2\pi\; i\sum_{r\text{ Residue}}\operatorname{Res}_{z=r} \frac {P(\ln z)}{(z-1)^2} =0 \ . $$ Now we let $R$ go to $+\infty$.

• The upper blue segment produces an integral of the shape $$\displaystyle \int_{0+\varepsilon}^\infty \frac{P(\ln x)}{(x-1)^2}\; dx\ , $$

• then the semicircle does not contribute, since for $z=R e^{it}$ the numerator contributes with $\ln^3 R+O(1)$, the denominator is $O(R^{-2})$, the whole fraction is $\sim R^{-2}\ln R$, and the contour is in $O(R)$,

• and finally the lower pink segment produces an integral of the shape $$\displaystyle \int_\infty^{0+\varepsilon} \frac{P(\ln x+2\pi i)}{(x-1)^2}\; dx\ , $$ if we succeed to push it beyond the holes.

Here i have to appologize for using $\ln$ for the real function $\log$, and also for the used branch of the logarithm, which is the reason for getting $\ln(x+2\pi i)$ by the monodromy of the logarithm when the contour comes back.

By the choice of $P$, the two integrals come together to build (up to sign) $$ \int_{0+\varepsilon}^\infty \frac{P(\ln x+2\pi i)-P(\ln x)}{(x-1)^2}\; dx = \int_{0+\varepsilon}^\infty \frac{x^2}{(x-1)^2}\; dx \ , $$ and this value comes then during "pushing" exactly from the holes in $0,1$. Let us compute the two contributions. Let $C(z_0,\varepsilon)$ the contour / the circle centered in $z_0\in\Bbb C$ with radius $\varepsilon>0$. Then using the substitution $z=z_0+\varepsilon e^{it}$, $t\in[0,2\pi]$, $$ \left| \int_{C(0,\varepsilon)} \frac {\ln^k z} {(z-1)^2}\; dz \right| \sim 2\pi\epsilon\cdot |\ln\varepsilon+O(1)|^k \in O(\varepsilon^{1/2}) $$ produces no contribution, and $$ \begin{aligned} \int_{C(1,\varepsilon)} \frac {\ln^k z} {(z-1)^2}\; dz &= \int_0^{2\pi} \frac {\displaystyle\ln^k(1+\varepsilon e^{it})} {\displaystyle(1+\varepsilon e^{it}-1)^2} \; i\varepsilon e^{it}\; dt \\ &= \int_0^{2\pi} \frac {\displaystyle \left( \frac 11\varepsilon e^{it} -\frac 12\varepsilon^2 e^{2it} +\frac 13\varepsilon^3 e^{3it} \pm\dots \right)^k} {\displaystyle \varepsilon^2 e^{2it}} \; i\varepsilon e^{it}\; dt \\ &\to \begin{cases} 2\pi i = a&\text{ for }k=1\ ,\\ 0&\text{ for }k\ge 2\ , \end{cases} \qquad \text{ when } \varepsilon\to0\ . \end{aligned} $$ The coefficient of $x$ in $P$ was $\frac a6$, so we get the contribution from $1$ in the form (well, computations were done up to sign) $$ \pm \frac 16a^2=\pm\frac 16(2\pi i)^2=\mp \color{blue}{\frac23\pi^2}\ . $$ (We are choosing the plus sign of course, the integral is built w.r.t. a positive function.)

$\color{red}\square$

Job done, but...

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This method applies mot-a-mot to the integrals $$ \int_0^\infty \frac{\ln^4 x}{(x-1)^2}\; dx\ ,\qquad \int_0^\infty \frac{\ln^6 x}{(x-1)^2}\; dx\ ,\qquad \int_0^\infty \frac{\ln^8 x}{(x-1)^2}\; dx\ ,\qquad \dots $$ in the following way:

• we build the corresponding Bernoulli polynomial $Q$ so that $Q(x+1)-Q(x)=(2n+1)x^{2n}$. An $a$-homogenized, $1/(2n+1)$-normed version of it delivers $P$ with $P(0)=0$, $P(x+a)-P(x)=x^{2n}$, and store the coefficient $c_1$ of $P(x)$ in $x^1$,
• we use the same contour of integration, the same argument applies,
• there is no contribution from the singularity in zero,
• the singularity in one contributes with $a=2\pi i$ again.

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Numerical experiments.

Numerically, i prefer to compute the half value, $$ \begin{aligned} J(2n) &= \int_1^\infty \frac{\ln^{2n} x}{(x-1)^2}\; dx \qquad\text{ Substitution: } y=\frac{1}{x},\ x=\frac{1}{y},\ dx=-\frac 1{y^2}\; dy \\ &= \int_0^1 \frac{\ln^{2n} y}{(1-y)^2}\; dy =\frac 12\int_0^\infty \frac{\ln^{2n} x}{(x-1)^2}\; dx \ . \end{aligned} $$ Then the expected value of the integral is numerically validated:

sage: for n in [2, 4, 6, 8, 10]:
....:     Q = bernoulli_polynomial(x, n+1)/(n+1)
....:     c1 = Q.coefficient( x^1 )
....:     print "n = %s" % n
....:     print "B(x, %s) / %s is %s" % (n+1, n+1, Q)
....:     print "Its coefficient in x^1 is %s" % c1
....:     f(x) = log(x)^n / (x-1)^2
....:     myintegral = numerical_integral( lambda x: f(x), (0,1) )[0]
....:     print "f(x) is %s" % (f(x))
....:     print "Integral of f on (0, 1) is ~ %f" % myintegral
....:     print "| c1 * (2 pi)^%2s / 2 |  is ~ %f" % (n, (abs(c1)*(2*pi)^n/2).n())
....:     print
....:     
n = 2
B(x, 3) / 3 is 1/3*x^3 - 1/2*x^2 + 1/6*x
Its coefficient in x^1 is 1/6
f(x) is log(x)^2/(x - 1)^2
Integral of f on (0, 1) is ~ 3.289868
| c1 * (2 pi)^ 2 / 2 |  is ~ 3.289868

n = 4
B(x, 5) / 5 is 1/5*x^5 - 1/2*x^4 + 1/3*x^3 - 1/30*x
Its coefficient in x^1 is -1/30
f(x) is log(x)^4/(x - 1)^2
Integral of f on (0, 1) is ~ 25.975758
| c1 * (2 pi)^ 4 / 2 |  is ~ 25.975758

n = 6
B(x, 7) / 7 is 1/7*x^7 - 1/2*x^6 + 1/2*x^5 - 1/6*x^3 + 1/42*x
Its coefficient in x^1 is 1/42
f(x) is log(x)^6/(x - 1)^2
Integral of f on (0, 1) is ~ 732.487004
| c1 * (2 pi)^ 6 / 2 |  is ~ 732.487005

n = 8
B(x, 9) / 9 is 1/9*x^9 - 1/2*x^8 + 2/3*x^7 - 7/15*x^5 + 2/9*x^3 - 1/30*x
Its coefficient in x^1 is -1/30
f(x) is log(x)^8/(x - 1)^2
Integral of f on (0, 1) is ~ 40484.398950
| c1 * (2 pi)^ 8 / 2 |  is ~ 40484.399002

n = 10
B(x, 11) / 11 is 1/11*x^11 - 1/2*x^10 + 5/6*x^9 - x^7 + x^5 - 1/2*x^3 + 5/66*x
Its coefficient in x^1 is 5/66
f(x) is log(x)^10/(x - 1)^2
Integral of f on (0, 1) is ~ 3632409.110395
| c1 * (2 pi)^10 / 2 |  is ~ 3632409.114224

sage: 

Also for some two bigger values:

sage: for n in [20, 30 ]:
....:     Q = bernoulli_polynomial(x, n+1)/(n+1)
....:     c1 = Q.coefficient( x^1 )
....:     print "n = %s" % n
....:     print "c1 is %s" % QQ(c1).factor()
....:     f(x) = log(x)^n / (x-1)^2
....:     myintegral = numerical_integral( lambda x: f(x), (0, 1), eps_abs=1e-4 )[0]
....:     print "f(x) is %s" % (f(x))
....:     print "Integral of f on (0, 1) is ~ %r" % myintegral
....:     print "| c1 * (2 pi)^%2s / 2 |  is ~ %r" % (n, (abs(c1)*(2*pi)^n/2).n())
....:     print
....:     
n = 20
c1 is -1 * 2^-1 * 3^-1 * 5^-1 * 11^-1 * 283 * 617
f(x) is log(x)^20/(x - 1)^2
Integral of f on (0, 1) is ~ 2.432904323980626e+18
| c1 * (2 pi)^20 / 2 |  is ~ 2.43290432907279e18

n = 30
c1 is 2^-1 * 3^-1 * 5 * 7^-1 * 11^-1 * 31^-1 * 1721 * 1001259881
f(x) is log(x)^30/(x - 1)^2
Integral of f on (0, 1) is ~ 2.6525284549431265e+32
| c1 * (2 pi)^30 / 2 |  is ~ 2.65252860059228e32

sage: 

Answer 2

I strongly agree with the comment by J.G.: the improper Riemann integral equals $$\int_{-\infty}^{+\infty}\frac{z^2 e^z}{(e^z-1)^2}\,dz = \int_{-\infty}^{+\infty}\left(\frac{z}{2\sinh\frac{z}{2}}\right)^2\,dz\stackrel{sym}{=}4\int_{0}^{+\infty}\frac{u^2}{\sinh^2 u}\,du$$ or, by integration by parts, $$ 8\int_{0}^{+\infty}u(\coth u-1)\,du=8\int_{0}^{+\infty}\left[u-\log(2\sinh u)\right]\,du = 8\int_{1}^{+\infty}\log\left(\frac{t}{t-1/t}\right)\frac{dt}{t}. $$ This can be shown to be equal to $4\,\zeta(2)=\frac{2\pi^2}{3}$ in many ways, for instance by reducing the last integral (via $t\mapsto 1/t$) to a multiple of

$$\int_{0}^{1}\frac{-\log(1-t^2)}{t}\,dt=\sum_{n\geq 1}\frac{1}{n}\int_{0}^{1}t^{2n-1}\,dt=\frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{12}.$$