In early 2020, I had an interview where I was asked to compute ∫xlogxdx.
A trick I learned from Jaynes:
∫xlogxdx=dda(∫xadx)|a=1=dda(1a+1xa+1)|a=1=(−1(a+1)2xa+1+1a+1xa+1logx)|a=1
=12x2logx−14x2
… or, if you want to be a smartass about it, write “1” in place of “a” and tell people you’re “differentiating by 1”: xlogx=dd1x1.
All you actually need to remember is xnlogx=ddnxn.
A trick I learned from Jaynes:
∫xlogxdx=dda(∫xadx)|a=1=dda(1a+1xa+1)|a=1=(−1(a+1)2xa+1+1a+1xa+1logx)|a=1
=12x2logx−14x2
… or, if you want to be a smartass about it, write “1” in place of “a” and tell people you’re “differentiating by 1”: xlogx=dd1x1.
All you actually need to remember is xnlogx=ddnxn.