Spence function

[Albert Chan]

(01-08-2021 01:28 AM)Albert Chan Wrote:  For k=1, g(t) is simple enough that Wolfram Alpha can proof it (after a few seconds)

I = ∫(ln((2+t)/(2-t)*(2t+1))/t, t = 0 .. 1) = pi^2/4

I just learned how to proof this, using Spence's function (dilogarithm)



Break-up Integral to parts, and integrate each:

∫(ln(1+t/2)/t - ln(1-t/2)/t + ln(1+2t)/t) dt
= (-Li2(-t/2)) - (-Li2(t/2)) + (-Li2(-2t)) + C

I = -Li2(-1/2) + Li2(1/2) - Li2(-2)
  = LI2(1/2) - (Li2(-2) + Li2(-1/2))

Identity:    Li2(z) + Li2(1-z) = pi^2/6 - ln(z)*ln(1-z)
Let z=1/2: Li2(1/2) * 2 = pi^2/6 - ln(1/2)^2 = pi^2/6 - ln(2)^2

Identity:    Li2(z) + Li2(1/z) = -pi^2/6 - ln(-z)^2/2
Let z=-2:   Li2(-2) + Li2(-1/2) = -pi^2/6 - ln(2)^2/2

I = (pi^2/6 - ln(2)^2)/2 - (-pi^2/6 - ln(2)^2/2) = pi^2/4

[Albert Chan]

(01-07-2021 09:37 PM)Valentin Albillo Wrote:  

The funny thing about this integral is that its value remains the same if you replace 2.021 by any positive real value !!

I think Spence's function (Dilogarithms) is the key to proof I = pi^2/4, for any k

After transformation, we have I = ∫(g(t) dt, t = 0 .. 1)

\(\exp(g(t)·k\,t) = \Large
{(1+(1+t)^k)·(1+({1\over t}+1)^k) \over (1+(1-t)^k)·(1+({1\over t}-1)^k)} =
{(1+(1+t)^k)·(t^k+(1+t)^k) \over (1+(1-t)^k)·(t^k+(1-t)^k)}\)

For now, assume k is positive integer → numerator is polynomial of t, with degree 2k.

Let W = exp((2n+1)/k*pi*i), n = 0 .. k-1. In other words, W = roots of x^k = -1

roots of numerator   , t = W-1, 1/(W-1)
roots of denominator, t = 1-W, 1/(W+1)

Consider just 1 of W (we have k of them), factor it out, scaled away k, and integrate:

\(\displaystyle \int _0^1 \ln\left( {(1-{t \over w-1})·(1-t(w-1)) \over
(1-{t \over 1-w})·(1-t(w+1)) }\right) {dt \over t} \normalsize
= -Li_2({1\over w-1}) + Li_2({1\over 1-w}) - Li_2(w-1) + Li_2(w+1) \)

I = average of all the k pieces, dropped imaginary parts (since I is real)

For example, for k = 7, these are the integral pieces.

>>> from mpmath import *
>>> Li2 = lambda x: polylog(2, x)
>>> k = 7
>>> W = [exp((2*n+1)/k*pi*1j) for n in range(k)]
>>> for w in W: print -Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(w+1)
...
(2.46740110027234 + 5.0951007958138j)
(2.46740110027234 + 2.21291211873152j)
(2.46740110027234 + 0.940187304494618j)
(2.46740110027234 + 1.22410719457408e-16j)
(2.46740110027234 - 0.940187304494618j)
(2.46740110027234 - 2.21291211873152j)
(2.46740110027234 - 5.0951007958138j)

All the real parts have the same size = pi^2/4

What happened if k is not integer ? Lets try k = 2.021
Amazingly, for any non-zero real k, real part of the integral piece still = pi^2/4

>>> w = exp(pi*1j/2.021)
>>> print -Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(w+1)
(2.46740110027234 + 1.85770404643526j)

Note: this only show how integral evaluated to the same value, for different k's
It is still not a proof, but much closer than what I had before ...

[Albert Chan]

Here is the proof for non-zero integer k, I = pi^2/4

Let A(w) = re(-Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(1+w))

Code:

  -Li2(1/(w-1)) - Li2(w-1)  +            Li2(1/(1-w))             + Li2(1+w)
= (pi*pi/6 + ln(1-w)^2/2)   + (-pi*pi/6 - ln(w-1)^2/2 - Li2(1-w)) + Li2(1+w)
= (ln(1-w)^2 - ln(w-1)^2)/2 + (Li2(1+w) - Li2(1-w))
= ln(-(1-w)^2) * (ln(1-w)-ln(w-1))/2 + (Li2(1+w) - Li2(1-w))

We like to show A(w) + A(-w) = pi^2/2, which implied I = average(A(W)) = pi^2/4
Using symmetry, we avoided calculations of Li2(x), since all will cancelled out. (*)

First factor, let w = c+s*i, where c=cos(x), s=sin(x), so that |w|=1

Code:

  -(1-w)^2 
= -(1-c-s*i)^2 
= -(1-c)^2     +   s^2   + 2*(1-c)*s*i
= -(1-2*c+c^2) + (1-c^2) + 2*(1-c)*s*i
= 2*(1-c)*c              + 2*(1-c)*s*i
= 2*(1-c)*w

Similar way, -(1+w)^2 = -2*(1+c)*w

Second factor, by definition, ln(x) = ln(|x|) + arg(x) * i

ln(x) - ln(-x) = (arg(x) - arg(-x)) * i = ±pi * i

Sign depends on phase angle. If we assume arg(w) > 0, we have

ln(1-w) - ln(-(1-w)) = -pi*i
ln(1+w) - ln(-(1+w)) = pi*i

re(ln(-(1-w)^2) * (ln(1-w) - ln(-(1-w)))) = arg(w) * pi
re(ln(-(1+w)^2) * (ln(1+w) - ln(-(1+w)))) = arg(-w) * -pi = (pi - arg(w)) * pi

A(w)  = arg(w)*pi/2          + re(Li2(1+w) - Li2(1-w))
A(-w) = (pi - arg(w))*pi/2 + re(Li2(1-w) - Li2(1+w))

→ A(w) + A(-w) = pi^2/2       QED

Note that k does not appear anywhere in the proof.
So, as long as we pair w with -w, even for non-integer k, above should hold.

---
(*) For odd k, w=-1 does not have anyone to pair with, but evaluation is simple.

A(w) = re(ln(-(1-w)^2) * (ln(1-w)-ln(w-1))/2 + (Li2(1+w) - Li2(1-w)))
A(-1) = pi^2/2 - re(Li(2))

Identity: Li2(z) + Li2(1-z) = pi^2/6 - ln(z) * ln(1-z)
Let z=-1: Li2(-1) + Li2(2) = pi^2/6 - ln(-1) * ln(2)

Li2(-1) = -pi^2/12
Li2(2) = (pi^2/6 - Li2(-1)) - ln(2)*pi*i = pi^2/4 - ln(2)*pi*i

A(-1) = p^2/2 - pi^2/4 = pi^2/4

To keep the symmetry, we can assign A(1) = pi^2/4

[C.Ret]

(01-11-2021 06:13 PM)Albert Chan Wrote:  I just learned how to proof this, using Spence's function (dilogarithm)



Break-up Integral to parts, and integrate each:

Wait a minute !

I just realize that I don't know how to integrate such an integral ; .

When the argument were any real number, I know how to get a rough numeric approximation using one of my useful and capable Hp calculators. Typing a short keystrokes code may help me to compute a value on my HP-15C.

For example,

Code:

g P/R f CLEAR PRGM  f LBL 2  1  -  CHS  g LN  X><Y  /  g RTN  g P/R
0  ENTER  .95  f Intyx 2
    flashing runnin9 ....
    -1.44063  CHS
     1.44063

From this computation carry on my venerable HP-15C, I deduce that Li2( 0.95) ~= 1.44063

Is that right ? If not, what I have done the wrong way ?


But, I suddenly discover that I have no way when argument is a complex (even with Flag 8 Set to complex mode). I even don’t know how to proceed with paper & pencil to integrate in complex plane.

How are mathematicians dealing with the computation of, for example :




I will thanks anyone here who will be kid enough to lead me to the right procedure

best regards.
C.Ret

[Albert Chan]

This post extended A(w) + A(-w) = pi^2/2, to a proof that A(w) = pi^2/4

Identity: Li2(z) + Li2(1-z) = pi^2/6 - ln(z)*ln(1-z)

Li2(1-w) = pi^2/6 - ln(w)*ln(1-w) - Li2(w)
Li2(1+w) = pi^2/6 - ln(-w)*ln(1+w) - Li2(-w)

Li2(1+w) - Li2(1-w) = ln(w)*ln(1-w) - ln(-w)*ln(1+w) + (Li2(w) - Li2(-w))

The reason for the conversion is because of this amazingly neat identity

\(\displaystyle\Re(Li_2(e^{iθ})) = \sum_{n=1}^∞ {\cos(nθ) \over n^2}
= {\pi^2\over6} - {\pi θ \over 2} + {θ^2 \over 4},\qquad 0 ≤ θ ≤ 2\pi\)

Applying formula for arg(w = e^(ix)) range of (-pi, pi], we have a sawtooth pattern:

Li2(w) - Li2(-w) = pi^2/4 - |arg(w)|*pi/2

From previous post, we have (1±w)^2 = ±2*(1-c)*w, where c = re(w).
Thus, arg(1±w) = arg(w)/2, with possible shifts.

Excluding w = 1, i.e. range(arg(w)) = (-pi,0) + (0,pi]
Let a = arg(w), s = sign(a) = ±1

arg(w) = a
arg(-w) = a - s*pi
arg(1-w) = a/2 - s*pi/2
arg(1+w) = a/2

ln(±w) is pure imaginary, this make pulling out real parts of terms easy.

re(ln(w)*ln(1-w) - ln(-w)*ln(1+w)) = (-a)*(a/2-s*pi/2) + (a-s*pi)*(a/2) = 0

A(w) = re(-Li2(1/(w-1)) + Li2(1/(1-w)) - Li2(w-1) + Li2(1+w))
        = |arg(w)|*pi/2 + re(Li2(1+w) - Li2(1-w))       // previous post, extended to all w
        = |arg(w)|*pi/2 + re(Li2(w) - Li2(-w))             // all log terms disappeared
        = |arg(w)|*pi/2 + pi^2/4 - |arg(w)|*pi/2
        = pi^2/4

Comment: This may be enough to show I = pi^2/4, for any non-zero real k.

[Albert Chan]

(01-12-2021 05:28 PM)C.Ret Wrote:  

(01-11-2021 06:13 PM)Albert Chan Wrote:  I just learned how to proof this, using Spence's function (dilogarithm)



Break-up Integral to parts, and integrate each:

Wait a minute !

I just realize that I don't know how to integrate such an integral ; .

Just move complex z inside the function. Let \(u = z\; x\;,\;du = z\;dx\)

\(\displaystyle Li_2(z) = - \int _0^1 {\ln(1-z\;x) \over x} dx\)

This substitution of variable give neat result. Example, let \(u = x^k\;,\;du = k\;x^{k-1}\;dx\)

\(\displaystyle -k \int _0^1 {\ln(1-x^k) \over x} dx
= -k \int _0^1 {\ln(1-u) \over x} {du \over k\;x^{k-1}}
= - \int _0^1 {\ln(1-u) \over u} du
= Li_2(1) = {\pi^2 \over 6} \)

I thought above formula had something to do with Valentin's puzzle ... but no luck.

[C.Ret]

(01-12-2021 07:49 PM)Albert Chan Wrote:  Just move complex z inside the function. Let \(u = z\; x\;,\;du = z\;dx\)

\(\displaystyle Li_2(z) = - \int _0^1 {\ln(1-z\;x) \over x} dx\)

This substitution of variable give neat result. Example, let \(u = x^k\;,\;du = k\;x^{k-1}\;dx\)

\(\displaystyle -k \int _0^1 {\ln(1-x^k) \over x} dx
= -k \int _0^1 {\ln(1-u) \over x} {du \over k\;x^{k-1}}
= - \int _0^1 {\ln(1-u) \over u} du
= Li_2(1) = {\pi^2 \over 6} \)

I thought above formula had something to do with Valentin's puzzle ... but no luck.

Thank you , for this quick response.

You exactly respond to my question. No more worries now about complex values in integration's limits.

Effectively, integrating from \(0\) to \(z\) on variable \(du\) give the same integration as from \(0\) to \(1\) on variable \(dx\) where \(x=\frac{u}{z}\)

You make my day. I learn something today. (I also discover that Latex is enable on this forum. Two nice discoveries in one post ).

As usual, a lot of nice people to meet here.
Thank you again.

[Albert Chan]

(01-12-2021 05:28 PM)C.Ret Wrote:  How are mathematicians dealing with the computation of, for example :

(01-12-2021 02:17 PM)Albert Chan Wrote:  Li2(-1) = -pi^2/12
Li2(2) = (pi^2/6 - Li2(-1)) - ln(2)*pi*i = pi^2/4 - ln(2)*pi*i

I was playing around, and tried integrating using Li2(2) as base.
0 to 2+3i → 0 to 2, then 2 to 2+3i

Turns out, we have to use the conjugate of the base.
Anyone knows how and when we apply the conjugate ?

XCas> b := pi^2/4 - ln(2)*pi*i                                   // Li2(2)
XCas> float(b)
2.46740110027 - 2.1775860903*i

XCas> conj(b) - i*int(ln(-1-x*i)/(2+x*i), x = 0. .. 3)    // let u = 2+x*i, du = i*dx
-0.280988055379 + 3.01725120637*i

XCas> -int(ln(1-(2+3i)*x)/x, x = 0. .. 1)                     // confirmed Li2(2+3i)
-0.280988055378 + 3.01725120637*i

FYI, picking a closer base speed up integration. (about 5X for above example)
The problem is how to make it right ...

[Albert Chan]

(01-12-2021 05:41 PM)Albert Chan Wrote:  re(ln(w)*ln(1-w) - ln(-w)*ln(1+w)) = (-a)*(a/2-s*pi/2) + (a-s*pi)*(a/2) = 0

Part of the prove for I = pi^2/4 involved proving above statement.

It may be more elegantly done using geometry


arg(w) = ∠COB = θ
arg(1+w) = ∠CAB = θ/2, since |w| = 1

arg(-w) = β = θ ± pi, sign = sign(-θ)
arg(1-w) = arg(-(1+w)) = β/2

re(ln(w)*ln(1-w) - ln(-w)*ln(1+w)) = (-θ)(β/2) + (β)(θ/2) = 0

[Albert Chan]

(01-12-2021 11:24 PM)Albert Chan Wrote:  I was playing around, and tried integrating using Li2(2) as base.
0 to 2+3i → 0 to 2, then 2 to 2+3i

Turns out, we have to use the conjugate of the base.
Anyone knows how and when we apply the conjugate ?

By experiments, I figured out how and when to apply conjugate, but don't know why ...
Consider only imaginery part. If Spence integral limits has 1 positive, but the other not, apply conjugate.

>>> from mpmath import *
>>> mp.pretty = True
>>> Li2 = lambda x: polylog(2, x)
>>> F = lambda a, b: quad(lambda x: -ln(1-x)/x, [a, b])

>>> Li2(2+3j)
(-0.28098805537806 + 3.01725120636941j)
>>> F(0, 2+3j)
(-0.28098805537806 + 3.01725120636941j)

>>> conj(Li2(2)) + F(2, 2+3j)                       # 2 is real, no positive imaginery part
(-0.280988055378061 + 3.01725120636941j)

Say, we have Li2(1-1j), how to correct it to Li2(2+3j) ?
We interpolate (1-1j), (2+3j) at the real axis.

x = x1 - y1*(x1-x2)/(y1-y2) = 1 - (-1)*(1-2)/(-1-3) = 1.25

>>> a, b, x = 1-j, 2+3j, 1.25
>>> conj(Li2(a) + F(a,x)) + F(x,b)                # == conj(Li2(x)) + F(x,b)
(-0.28098805537806 + 3.01725120636941j)

---

Without a close reference, we could use these identities to convert argument.

#1: Li2(z) + Li2(z/(z-1)) = -ln(1-z)^2/2
#2: Li2(z) + Li2(1-z) =  pi^2/6 - ln(z) * ln(1-z)
#3: Li2(z) + Li2(1/z) = -pi^2/6 - ln(-z)^2/2

We have to be careful with #3 to get Li2(z), if Li2(z) is real, but Li2(1/z) is complex.
This may be why https://mathworld.wolfram.com/Dilogarithm.html excluded it.

>>> z = 0.95
>>> Li2(z) , -pi**2/6 - ln(-z)**2/2 - Li2(1/z)       # bad
(1.44063379697004, (1.44063379697004 + 0.322285273652695j))
>>> z = 1/0.95
>>> Li2(z) , -pi**2/6 - ln(-z)**2/2 - Li2(1/z)       # ok
((1.84791883570185 - 0.161142636826347j), (1.84791883570185 - 0.161142636826347j))

For b = 2+3j, it is ok to apply. 1/b is tiny enough we can sum, instead of integrate.

>>> x = 1/b
>>> s = sum(x**k/(k*k) for k in range(24,0,-1)) # Li2(x)
>>> -pi**2/6 - ln(-b)**2/2 - s
(-0.280988055378061 + 3.01725120636941j)      # Li2(b)

The other 2 identities seems safe ...

#1: Li2(2+3j) = -ln(-1-3j)^2/2 - Li2(1.1-.3j)
#2: Li2(1.1-.3j) = pi^2/6 - ln(1.1-.3j)*ln(-.1+.3j) - Li2(-.1+.3j)

>>> x = -.1 + .3j
>>> s = sum(x**k/(k*k) for k in range(24,0,-1)) # Li2(x)
>>> s = pi**2/6 - ln(1.1-.3j)*ln(-.1+.3j) - s        # Li2(1-x)
>>> -ln(-1-3j)**2/2 - s                                      # Li2(b)
(-0.28098805537806 + 3.01725120636941j)

[Albert Chan]

(01-14-2021 01:55 PM)Albert Chan Wrote:  >>> a, b, x = 1-j, 2+3j, 1.25
>>> conj(Li2(a) + F(a,x)) + F(x,b)                # == conj(Li2(x)) + F(x,b)
(-0.28098805537806 + 3.01725120636941j)

This is how WolframAlpha interpret F(a, b):

∫(-ln(1-x)/x, x = 1-I .. 2+3I)
= (Li2(2+3I) - conj(Li2(1.25))) + (Li2(1.25) - Li2(1-I))
= (Li2(2+3I) - Li2(1-I)) + (Li2(1.25) - conj(Li2(1.25)))
= (Li2(2+3I) - Li2(1-I)) + 2I * im(Li2(1.25))

Do not calculate F(a, b) directly !
Because of the discontinuity crossing the real axis, it it very hard to evaluate accurately.

Since we are interested with Li2(2+3I), interpolation of integral limits is not necessary.
Any real x will do. Preferably Li2(x) is also real, so no need to worry about conjugates.

>>> pi*pi/6 + F(1, 2+3j)       # x = 1
(-0.280988055378061 + 3.01725120636941j)

This simple shift doubled the speed of integration, compared against F(0, 2+3j)

[Albert Chan]

Chaos in Numberland: The secret life of continued fractions, by John D. Barrow

In this great article, there was an identity that we can prove, using Spence function.
(section: What is chaos ?, equation 36, where h = Kolmogorov, or metric, entropy of the mapping)



Do integration by parts, where \(u = \ln(x), v = \ln(1+x) \)

\(\displaystyle \int {\ln(x) \over 1+x}\,dx
= \int \ln(x)\,d(\ln(1+x))
= \ln(x)\ln(1+x)\;+ \int {-\ln(1+x) \over x}\,dx
\)

Adding limit of integration, x = 0 to 1, last term = Li2(-1)

At x=1, uv = ln(1) ln(2) = 0
At x=0, uv = ln(0) ln(1) = -∞ * 0 = ?

Note: At x=0, v = ln(1+x) = 0 , v' = 1/(1+x) = 1

\(\displaystyle \lim_{x \to 0} {u \over v^{-1}}
= \lim_{x \to 0} {u' \over -v^{-2}\;v'}
= \lim_{x \to 0} {-v^2 \over x}
= \lim_{x \to 0} {-2v\,v' \over 1}
= 0 \)

Or, we can avoid repeated applications of L'Hospital rule, by letting x = 1/n

\(\displaystyle \lim_{ x \to 0}\,(uv)
= \lim_{n \to ∞} \left[ \left({-\ln(n) \over n}\right)
\ln\left(\left(1+{1\over n}\right)^n\right) \right]
= (-0) (1) = 0 \)

For n → ∞, n ≫ ln(n) ; e = limit((1+1/n)^n, n=inf), by definition

Either way, uv term disappered, and we are left with last term, Li2(-1) = -pi^2/12

\(\displaystyle h
= {-2 \over \ln(2)} \int_0^1 {\ln(x) \over 1+x}\,dx
= {-2 \over \ln(2)} {-\pi^2 \over 12}
= {\pi^2 \over 6\ln(2)} \)

[Albert Chan]

(03-24-2017 05:12 PM)emece67 Wrote:  \[\int_{-1}^\infty{\ln(x+1)\over x^2+1}dx={3\pi\ln(2)\over8}\approx0.8165947838638507989377583368391052\ldots\]

I was curious to the proof of this. Turns out, it can be solved using Spence function.
Break up integral to three parts, then normalized to unit interval.

\(\displaystyle \int_{-1}^0 + \int_0^1 + \int_1^∞
= \int_0^1 \left[ {\ln(1-x)\over 1+x^2}
+ {\ln(1+x)\over 1+x^2}
+ {\ln(1+{1\over x})\over 1+x^2}
\right]dx
= \int_0^1 \left[ {\ln(1-x)\over 1+x^2}
+ {2\ln(1+x)\over 1+x^2}
- {\ln(x)\over 1+x^2}
\right]dx \)

(01-12-2021 07:49 PM)Albert Chan Wrote:  Just move complex z inside the function. Let \(u = z\; x\;,\;du = z\;dx\)

\(\displaystyle Li_2(z) = - \int _0^1 {\ln(1-z\;x) \over x} dx\)

What is interesting is the first term. To match above form, we integrate from the right.

\(\displaystyle I_1
=\int_0^1 {\ln(1-x)\over 1+x^2} dx
= \int_0^1 {\ln(x)\;dx\over 1+(1-x)^2}\)

Do integration by part, u=-ln(x), v=atan(1-x), ∫u dv = u*v - ∫v du

\(\displaystyle I_1
= -\ln(x)\;atan(1-x)\bigg|_0^1\; +\int_0^1 {atan(1-x) \over x}\;dx \)

We have -∞ + ∞ !!!
But, if we left it alone for a while, infinity cancelled

From another thread about mnemonics:

(08-22-2020 04:10 PM)Albert Chan Wrote:  

Code:

erf (-x) = - erf(x)    → erf (ix) = i erfi(x)
tan (-x) = - tan(x)    → tan (ix) = i tanh(x)
tanh(-x) = - tanh(x)   → tanh(ix) = i tan(x)

tan(x) = i tanh(-xi)                         → atan(x) = i atanh(-xi)
atanh(x) = 1/2*ln((1+x)/(1-x))       → atan(x) = i/2 * ln((1-xi)/(1+xi)) = i/2 * ln((i+x)/(i-x))

\(\displaystyle \int_0^1 {atan(1-x) \over x}\;dx
= \int_0^1 {{i\over2} \ln\left({i+(1-x) \over i-(1-x)}\right)\over x}\;dx
= {i\over2} \int_0^1 \left[
{\ln\left({1+i \over -1+i}\right)\over x}
+ {\ln\left(1-{x \over 1+i}\right)\over x}
- {\ln\left(1-{x \over 1-i}\right)\over x}
\right]\;dx\)

i/2 * ln((1+i)/(-1+i)) = i/2 * ln(-i) = i/2 * (-i*pi/2) = pi/4
The other 2 terms have Li2 form, we are ready for I1 now.

\(\displaystyle I_1
= \ln(x)\left({\pi\over4}-atan(1-x)\right)\bigg|_0^1\; + {i\over2}
\left( Li_2({1\over1-i}) - Li_2({1\over1+i})\right)\)

Using L'Hospital Rule, first term is zero.
I1 = i/2*(Li2(1/(1-i)) - Li2(1/(1+i))) = pi*log(2)/8-C ≈ -0.643767

The other terms are done similarly ...

[Albert Chan]

(04-04-2021 10:57 PM)Albert Chan Wrote:  The other terms are done similarly ...

My mistake, only last term can be done similar way.

Quote:\(\displaystyle \int_{-1}^0 + \int_0^1 + \int_1^∞
= \int_0^1 \left[ {\ln(1-x)\over 1+x^2}
+ {\ln(1+x)\over 1+x^2}
+ {\ln(1+{1\over x})\over 1+x^2}
\right]dx
= \int_0^1 \left[ {\ln(1-x)\over 1+x^2}
+ {2\ln(1+x)\over 1+x^2}
- {\ln(x)\over 1+x^2}
\right]dx \)

To save time, I confirm my math in XCas (simplify Li2 mess using Wolfram Alpha)

I3 = int(-ln(x)/(1+x*x), x=0..1.)
    = preval(-ln(x)*atan(x),0,1) + int(atan(x)/x, x=0..1.)
    = i/2 * int(ln(1-x*i)/x - ln(1+x*i)/x, x=0..1.)
    = i/2 * (Li2(-i)-Li2(i))
    = C                               // Catalan's constant ≈ 0.915966

Using indefinite formula (confirmed by taking the derivative to match):

\(\displaystyle \int {\ln(x+a) \over x+b}\;dx =
Li_2\left( {x+a \over a-b} \right)
+ \ln(x+a) \ln\left({x+b \over b-a} \right)\)

I2 = 2 * int(ln(1+x)/(1+x^2), x=0..1)
    = i * int(ln(1+x)/(x+i) - ln(1+x)/(x-i), x=0..1)
    = i * preval(F(x), 0, 1)

F(x) = Li2((1+x)/(1-i)) - Li2((1+x)/(1+i)) + log(1+x)*(log((i+x)/(i-1)) - log((i-x)/(i+1)))       (*)
F(1) = Li2(1+i) - Li2(1-i) - i*log(2)*pi = 2i*C - i/2*log(2)*pi
F(0) = Li2(1/(1-i)) - Li2(1/(1+i)) = 2i*C - i/4*log(2)*pi

I2 = i*(F(1) - F(0)) = 1/4*log(2)*pi

I1 = i/2*(Li2(1/(1-i)) - Li2(1/(1+i))) = 1/8*log(2)*pi - C, from previous post

\(\displaystyle\int_{-1}^∞{\ln(x+1)\over 1+x^2}dx = I = I_1 + I_2 + I_3 = {3\over8}\ln(2)\,\pi\)

---

(*) difference of logs may not be combined, into log(i*(x+i)/(x-i))
Example:

\(\displaystyle \int {\ln(-x-a) \over x+b}\;dx =
Li_2\left( {x+a \over a-b} \right)
+ \ln(-x-a) \ln\left({x+b \over b-a} \right)\)

Difference is *NOT* \(\displaystyle \int {\ln(-1) \over x+b}\;dx = i\pi\ln(x+b)\)

For the same reason, log(x*y) may not be splitted, to log(x) + log(y)
Previous post had this bug (which turns out not to matter, trying to match Li2 form)

Now that we have F(x), we can confirm again I1 is correct.

I1 = i/2*(F(0) - F(-1)) = i/2*F(0) = 1/8*log(2)*pi - C

[Albert Chan]

I = i/2*(F(∞)-F(-1)) = i/2*F(∞) = 3/8*log(2)*pi

→ F(∞) = (-3i/4)*log(2)*pi ≈ -1.63319*i

>>> Li2 = lambda x: polylog(2,x)
>>> F1 = lambda x: Li2((1+x)/(1-j)) - Li2((1+x)/(1+j))
>>> F2 = lambda x: ln(1+x)*((ln((j+x)/(j-1)) - ln((j-x)/(j+1))))
>>> F = lambda x: F1(x) + F2(x)
>>> print F(1e18)
(0.0 - 1.6331895677277j)

Numerically, numbers agreed, suggesting F(x) is indeed correct.
We may be able to proof this:

Examining F1(x), Li2 arguments are complex conjugate of each other.
Conjugate symmetry applied to polylog functions, Li_p(z) = Σ(z^k/k^p, p=1..inf)

With Li2(conj(z)) = conj(Li2(z)), F1 is purely imaginery.

F1 = 2*im(Li2((1+x)/(1-j))) * i

(01-14-2021 01:55 PM)Albert Chan Wrote:  #3: Li2(z) + Li2(1/z) = -pi^2/6 - ln(-z)^2/2

F1/i
= 2*im(Li2(z))                       -- z = (1+x)/(1-j) = (1+x)/2 + (1+x)/2*i
≈ -im(ln(-z)^2)                     -- apply #3. when z is big, Li2(1/z) ≈ 0
= -im((ln(abs(z)) - 3/4*pi*i)^2)
= 3/2*pi*ln(abs(z))
= 3/2*pi*ln((1+x)/sqrt(2))
= 3/2*pi*ln(1+x) - 3/4*ln(2)*pi

F2/ln(1+x) = ln((i+x)/(i-1)) - ln((i-x)/(i+1))

RHS, arguments inside log are also conjugate of each other.
As expected, F2 is also purely imaginery.

Cas> simplify(conj((i+x)/(i-1)) - (i-x)/(i+1))       → 0

F2/ln(1+x)/i
= 2*arg((i+x)/(i-1))
= 2*arg((1-x) - (1+x)*i)
≈ 2*arg(-x - x*i)                   -- when x grows big, +1 has little effect
= -3/2*pi

F(∞) = limit(F1+F2, x=∞) = -3/4*ln(2)*pi*i

→ I = i/2*F(∞) = 3/8*ln(2)*pi

---

An efficient way to integrate, for \(\large\int_{-1}^x\)

>>> I = lambda x: ln(1+x)*atan2(1+x,1-x) - im(Li2((1+x)*(.5+.5j)))
>>> for x in range(5): print x, I(x)
...
0       -0.643767332889269
1       -0.371569071601318
2       -0.0888721439284095
3       0.0860509375677831
4       0.200606431044641

>>> I(1e18)
mpf('0.81659478386384876')

[Albert Chan]

Very beautiful ! We can proof I the same way, just different integral limits.

I = int(ln(1+x)/(1+x²), x=-1 .. ∞)       // x=tan(θ), dx = sec(θ)^2 dθ
  = int(ln(1+tan(θ)), θ=-pi/4 .. pi/2)
  = int(ln(√2), θ=-pi/4 .. pi/2) + int(ln(cos(θ-pi/4)) - ln(cos(θ)), θ=-pi/4 .. pi/2)

When we fold the second term, it disappeared !

∫(f(x), x=a..b) = ∫(f(x) + f(a+b-x), x=a..b) / 2

int(ln(cos(θ-pi/4)) - ln(cos(θ)) + ln(cos(-θ)) - ln(cos(pi/4-θ)), θ=-pi/4 .. pi/2) / 2 = 0

All is left is the first term, I = 3/8*ln(2)*pi

[Albert Chan]

(04-11-2021 03:22 AM)Albert Chan Wrote:  I = int(ln(1+x)/(1+x²), x=-1 .. ∞)       // x=tan(θ), dx = sec(θ)^2 dθ
  = int(ln(1+tan(θ)), θ=-pi/4 .. pi/2)
  = int(ln(√2), θ=-pi/4 .. pi/2) + int(ln(cos(θ-pi/4)) - ln(cos(θ)), θ=-pi/4 .. pi/2)

Simplify steps, tan(θ) = sin(θ)/cos(θ), then cos(θ) + sin(θ) = √(2)*cos(θ-pi/4) can all be skipped.
Folding tan(θ) integrand directly is easier.

I = ∫(ln(1+tan(θ)), θ=-pi/4 .. pi/2)
2*I = ∫(ln(1+tan(α)) + ln(1+tan(β)), α=-pi/4 .. pi/2)                   // where α = θ, β = pi/4 - α
      = ∫(ln(1 + tan(α) + tan(β) + tan(α)*tan(β)), α=-pi/4 .. pi/2)

tan(α+β) = (tan(α) + tan(β)) / (1 - tan(α)*tan(β)) = tan(pi/4) = 1

1 + tan(α) + tan(β) + tan(α)*tan(β) = 1 + (1 - tan(α)*tan(β)) + tan(α)*tan(β) = 2

2*I = ∫(ln(2), α=-pi/4 .. pi/2)

I = ln(2) * (pi/2 + pi/4) / 2 = 3/8*ln(2)*pi

[Albert Chan]

From [VA] SRC #010 - Pi Day 2022 Special

(03-17-2022 02:54 PM)Albert Chan Wrote:  \(\displaystyle \ln(\pi) = \int_0^{1/2} \left(
\frac{1}{y} + \frac{\pi (2y-1)}{\tan(\pi y)}
\right) dy\)

Wolfram Alpha proved this, with closed-form anti-derivative !

Only the midde term is troublesome. Lets get integral without middle term.

∫(1/y - pi/tan(pi*y) dy) = ln(y) - ln(sin(pi*y)) + C1 = -ln(sinc(pi*y)) + C2
∫(1/y - pi/tan(pi*y), y=0..1/2) = preval(-ln(sinc(pi*y)), 0, 1/2, y) = ln(pi) - ln(2)

For middle term integral, let x = pi*y, dx = pi dy

∫(2*pi*y / tan(pi*y), y=0..1/2) = 2/pi * ∫(x/tan(x), x=0 .. pi/2)

This reduced the proof to ∫(x/tan(x), x=0 .. pi/2) = pi/2*ln(2) ≈ 1.0888

∫(x/tan(x) dx) = ∫(x*i/tanh(x*i) dx) = ∫(z/tanh(z) dz) / i, where z=x*i

f = z/tanh(z) = z * cosh(z)/sinh(z) = z * (1+u)/(1-u), where u = e^(-2z)

To get (1-u) denominator, lets try ln(1-u)'

ln(1-u)' = 1/(1-u) * (1-u)' = 2u/(1-u)
(z*ln(1-u))' = ln(1-u) + z * ((1+u)-(1-u)) / (1-u) = ln(1-u) + f - z

Li2(u)' = -ln(1-u)/u * u' = 2*ln(1-u)

f = (z*ln(1-u))' + z - Li2(u)'/2
F = z*ln(1-u) + z^2/2 - Li2(u)/2       // = ∫(z/tanh(z) dz)

F(pi/2*i) = i*(pi/2*ln(2)) + (pi/2*i)^2/2 - Li2(-1)/2       // Li2(-1) = -pi^2/12
             = i*(pi/2*ln(2)) - pi^2/8 + pi^2/24

F(0*i) = limit(z*ln(1-exp(-2z)), z=0) + 0 - Li2(1)/2       // Li2(1) = ζ(2) = pi^2/6
         = 0 - pi^2/12

∫(x/tan(x), x=0..pi/2) = (F(pi/2*i) - F(0*i)) / i = pi/2*ln(2)       QED

---

It would be nice if F(0) = 0, avoiding calculation of limits.

For z = 0*i to pi/2*i, we have log(e^(2z)) = 2z. Let U = e^(2z):

f = z*(1+u)/(1-u) = -z*(1+U)/(1-U)

Li2(1-U)'
= -ln(1-(1-U)) / (1-U) * (1-U)'
= ln(U) / (1-U) * (2U)
= ln(U) / (1-U) * ((1+U) - (1-U))
= 2z*(1+U)/(1-U) - 2z         // NOTE: -pi/2 < im(z) ≤ pi/2
= -2f - 2z

f = -z - Li2(1-U)'/2
F = -z^2/2 - Li2(1-U)/2       // = ∫(t/tanh(t), t = 0 .. z), -pi/2 < im(z) ≤ pi/2

F(pi/2*i) = -(pi/2*i)^2/2 - Li2(2)/2       // Li2(2) = pi^2/4 - i*(pi*ln(2))

∫(x/tan(x), x=0..pi/2) = F(pi/2*i)/i = pi/2*ln(2)       QED