Re: Help evaluating a logarithmic integral

On Thu 8 Jun, 2023 at 08:31, Alessandro Languasco wrote:

Dear Kevin,

I read your question on the pari-list.

I think that the problem is that the function abs(x^v - (1 - x)^v) has a zero at x=1/2 and I suppose that
this creates some problem to the numerical integration of intnum. This is different form the other
case you wrote since x^v+(1-x)^v has no zeros for x into (0,1).

So I would try to verify first that the integral between (0,1/2) is convergent, and then
I would try to to split the integration range in two subintervals and use
intnum first onto (0,1/2) and then onto (1/2,1) .

You could even go as far as taking the double of the integral on (0,1/2).

Loïc

Best regards
Alessandro

> On 8 Jun 2023, at 03:10, kevin lucas <lucaskevin296@gmail.com> wrote:
>
> I have been attempting to investigate the function
> f(v) = intnumgauss(x=0,1, log(abs(x^v - (1 - x)^v))/x)
> for various positive values of v.
> I cannot use the standard intnum because I get a domain error for log.
> Even running on ~1000 precision I can't seem to get stable values. For example:
>
> ? f(2)
> %2 = -2.4225110046071715036046951871135247939
> ? \p 100
> realprecision = 115 significant digits (100 digits displayed)
> ? f(2)
> %3 = -2.448045467855465962637520950271434461626906750839030270690427142653795442332270916763571861586211446
> ? \p 500
> realprecision = 500 significant digits
> ? f(2)
> %4 = -2.46236623125534737591438595461875712342066622953365858291276735021494398975080267620885683763596204514917830504241706038254418887945208437689348318484881426720
> 6369967589343063020562790497652612860415817451002780032725265928653909391217140691580623706469465927763263097639596472178685417890225073853769466120537250499583029234
> 0105930878210069637749954513919387562716196350765008607089175830388519925771160765046638359503537136422685591455325755980613085912846141203505510512236119034080952188
> 143076548
>
> I have several questions:
> A. Why does intnum not work for this function? The similar integral g(v)=intnum(x=0,1,log(x^v+(1-x)^v)/x) is evaluated just fine, and agrees with intnumgauss.
> B. How should I best compute f? Suppose I wanted many accurate digits for several special values, to try and recognize them.
> C. Should I even be trusting intnumgauss for such an integrand?
>
>