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1 = 2 :)
\[\int\limits_{0}^{\infty} x ^{n} e ^{-x} dx\]
x^?
x to the n times e to the negative x. n is an integer
without knowing "n" this could be IBPed for a while ... at least til x^n derives to 0
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the n isn't important it can stay a constant. i need the integral with respect to x
\[n!\]
or maybe we use e^-x as the deriving factor and play with x^n till we get something that can be doubled
take my word for it
the n cant stay a constant, it is a power attached to "x"; which changes by the nature of the process i believe
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