deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 821163c..c3afa89 100644  --- a/untitled.tex  +++ b/untitled.tex 
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\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= -\Gamma(z) \\  \int_0^\infty x^{z-1}e^{-x}(-x+z-1) dx= -\Gamma(z) & \int_0^\infty x^{z-1}e^{-x} (x+x^2+(-1+z)^2-2xz)dx=? (x+x^2+(-1+z)^2-2xz)dx=\Gamma(z)  \\ \hline   \end{array}  \end{equation}  Above we have used (or discovered depending on your point of view!)  that $(z-1)\Gamma(z)-\Gamma(z+1)=-\Gamma(z)$. Then reused this again an used/discovered that $(1-2z)\Gamma(z+1)+\Gamma(z+2)+(z-1)^2\Gamma(z)=\Gamma(z)$...  Altohugh if one knows the inner workings of the gamma function this is perhaps obvious, we could potentially using this method, discover identities for horrendously complicated functions without knowing what they were... I will demonstrate below later on.  This derivative has explicity operated on functions of $x$, which makes it still in some sense "with respect to x". One could imagine multiple operations on multidimensional functions...