TITLE: Multiprecision numerical evaluation of integral: Sage vs. PARI/GP vs. mpmath
QUESTION [9 upvotes]: I am trying to compute thousands of integrals of the below type, that comes up in a conformal mapping problem, to as many accurate digits as possible (preferably 50+):
$$
\int_{-1}^1\textrm{d}t \frac{\mathcal{Re}\{\log[(\cos{(\pi/130)} - t)]\}}{\sqrt{1 - t^2}}
$$
The results from PARI/GP, Sage and Python's mpmath library respectively are:
-2.1770705767584673426214016567105099553,
(-2.1775860588840983, 1.2746272565903925e-05),
-2.1774410877577223893132496923831596284

Clearly $-2.177$ is correct, but what's the best way to find a more accurate answer, accurate to 50+ digits? 
I've tried splitting intervals from $[-1, \tau] \cup [\tau, 1]$, where $\tau = \cos{(\pi/130)}$; that doesn't improve things but actually makes it worse. I am working at much higher decimal precision than 3.
UPDATE: Following the suggestion of IgorRivin, the below is the Mathematica attempt:
tau = N[Cos[Pi/130], 50];
epsilon = 2^-10;
limit = N[ArcCos[1 - epsilon], 50];
T1 = NIntegrate[Re[Log[(tau - t)]]/(Sqrt[1 - t^2]), {t, -1, 1 - epsilon}, 
               WorkingPrecision -> 50, AccuracyGoal -> 50]
T2 = NIntegrate[Re[Log[tau - Cos[theta]]], {theta, 0, limit}, 
                WorkingPrecision -> 50, AccuracyGoal -> 50]
answer = T1 + T2

gives the results:
Out[1]= -1.7968036050143567231750633164742621583459497767361

During evaluation of In[881]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed 
accuracy after 9 recursive bisections in theta near 
{theta} = {0.024170580099064656300274166159163078224443688377727}. 
NIntegrate obtained -0.38078136551592029350719560689304843811203900465893 
and 6.9050153103011684224977203192777088512613553501207`50.*^-7 for the 
integral and error estimates. >>

Out[2]= -0.38078136551592029350719560689304843811203900465893

Out[3]= -2.1775849705302770166822589233673105964579887813950

As may be seen the error is still in the seventh decimal place; increasing the working precision or accuracy does not really improve things.
UPDATE: The integral is exactly soluble; the result is given in my comment to the accepted answer of Emil Jeřábek, and compares well to the exact value.

REPLY [10 votes]: Expanding my comment above: putting $\alpha=\pi/130$, the integral equals
$$\int_0^\pi\log\left|\cos\alpha-\cos\theta\right|d\theta=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\int_{-\alpha}^{\pi-\alpha}\log\left|\theta\sin\alpha\right|d\theta$$
$${}=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\pi\log\sin\alpha+\alpha(\log\alpha-1)+(\pi-\alpha)(\log(\pi-\alpha)-1).$$
The new integrand is regular on $[0,\pi]$, hence it has a better chance to be accurately approximated by numerical integration. I leave it to someone knowledgeable with such tools to try it.
In case it helps with cancellation errors, one can further write
$$\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta=\int_{-\alpha}^{\pi-\alpha}\log\left(\frac{\sin\theta}\theta+\frac{1-\cos\theta}\theta\cot\alpha\right)d\theta.$$