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\hline   \int_0^\pi sin(x)dx =2 & \int_0^\pi xcos(x)dx = -2 \\  \int_0^\infty e^{-x} dx =1 & \int_0^\infty -xe^{-x} dx = -1 \\  \int_0^\infty x^{z-1}e^{-x} dx= \Gamma(z) & \int_0^\infty x^{z-1}e^{-x}(-x+z-1) dx= (z-1)\Gamma(z)-\Gamma(z+1)=-\Gamma(z) -\Gamma(z)  \\ \hline   \end{array}  \end{equation}  Above we have used that $(z-1)\Gamma(z)-\Gamma(z+1)=-\Gamma(z)$  Contradictory examples to this are  \begin{equation}