Spence function

[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