Interesting find in integrating $f(\text{x})=e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right).$ Why?

Question

Interesting find in integrating $f(\text{x})=e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right).$ Why?

It looks like for

$$f(\text{x$\_$})=e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right),$$

$$-\frac{1}{2}\text{Re}\sum _{n=0}^{\infty } (f n) \int_0^{\infty } f t \, dt\approx 2 \text{Re} \int_0^{\infty } \frac{f (i t)}{\exp (2 \pi t)-1} \, dt.$$

$$ -\frac{1}{2} \left(\text{Re} \sum _{n=0}^{1512000} (f n) \int_0^{\infty } f t \, dt\right)=2 \left(\text{Re} \int_0^{\infty } \frac{f (i t)}{\exp (2 \pi t)-1} \, dt\right)\pm\frac{7}{10^{12}}.$$ Because in Mathematica,

-Re[NSum[f[n], {n, 0, 1512000}, Method -> "AlternatingSigns", 
     WorkingPrecision -> 20] NIntegrate[(f[t]), {t, 0, Infinity}, 
     WorkingPrecision -> 15]]/2

gives

-0.00664781420701289.

While

2 Re[NIntegrate[f[I t]/(Exp[2 \[Pi] t] - 1), {t, 0, \[Infinity]}, 
   WorkingPrecision -> 15]]

gives

-0.00664781419931721.