$\int_0^\infty e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) dx$

Posted 1 year ago
194862 Views
|
6 Replies
|
2 Total Likes
|

$\int_0^\infty e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) dx$

f[x_] = E^(I*Pi*x)*(1 - (x + 1)^(1/(x + 1)));
g[x_] = x^(1/x); u := t/(1 - t);

sub = Im[NIntegrate[(f[(-I t)] - f[( I t)])/(Exp[2 Pi t] - 1), {t,
0, Infinity}, WorkingPrecision -> 100]]

0.1170836031505383167089899122239912286901483986967757585888318959258587743002\
7817712246477316693025869

m = NSum[f[( t)] , {t, 0, Infinity}, WorkingPrecision -> 100,
Method -> "AlternatingSigns"]

0.1878596424620671202485179340542732300559030949001387861720046840894772315646\
6021370329665443217278

m - sub

0.0707760393115288035395280218302820013657546962033630275831727881636184572643\
8203658083188126524252

Is the same as

{Re[NIntegrate[f[t], {t, 0, Infinity}]], and,
Re[NIntegrate[f[t], {t, 0, Infinity I}, WorkingPrecision -> 100]]}

NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral estimate, but the integral might be divergent.

NIntegrate::deodiv: DoubleExponentialOscillatory returns a finite integral estimate, but the integral might be divergent.

{0.070776, and, \
0.0707760393115288035395280218302820013657546962033630275831727881636184572643\
8203658083188126617723821}


To be continued.

6 Replies
Sort By:
Posted 1 year ago
Posted 1 year ago
 I won't bring the subject of numerical computation into this discussion, but I spent several years learning to compute the digits of $\int_0^\infty{e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right)}dx.$ You can read about my adventure at How to calculate the digits of the MKB constant. If you like numeric computations, of much interest is the story of how I came across this integral, by investigating $\sum _{x=0}^{\infty } e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right)$ since the 1990s, at Try to beat these MRB constant records!
Posted 1 year ago
 So, I could conclude that The following "proof of it is from combining andWhile Something similar is shown: That looks like it proves to me. But I don't see any proof for .
Posted 1 year ago
Posted 1 year ago
Posted 1 year ago
 MRB=$\sum _{x=0}^{\infty } e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right)$ vs M2= $\int_0^\infty e^{i \pi x} \left(1-(x+1)^{\frac{1}{x+1}}\right) dx$ in proper integralsSee this notebook. I got an interesting co-answer here.