Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 2 I have stumbled into a issue with gamma function, I will show the approach first: $$Gamma(x)=int_0^inftyt^x-1e^-tdt$$ subtituting $t=iu^2$ gives: $$Gamma(x)=2int_0^infty(iu^2)^x-1e^-iu^2iudurightarrowfracGamma(x)2i^x=int_0^inftyu^2x-1e^-iu^2du$$ Doing the same thing using $t=-iu^2,$results in$$fracGamma(x)-2i^x=int_0^inftyu^2x-1e^iu^2du$$ Now, summing those two and using that $i^x=e^fracipi2x ,$gives $$fracGamma(x)2(e^fracipi2x+e^frac-ipi2x)=int_0^inftyu^2x-1(e^iu^2+e^-iu^2)du$$ which is just $$fracGamma(x)2cos(fracpi2x)=int_0^inftyu^2x-1cos(u^2)du$$ plugging $x=-frac12$ we get that$$int_0^inftyfraccos(x^2)x^2dx=-sqrtfracpi2$$ Well, obviously this integral diverges... But if Instead of summing we subtract we get that $$fracGamma(x)2sin(fracpi2x)=int_0^inftyu^2x-1sin(u^2)du$$ simmilarly with $$x=-frac12 rightarrow int_0^inftyfracsin(x^2)x^2dx=sqrtfracpi2$$ So its not that completely garbage. Now my question is, w...
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