Evaluating a Pretty Big Integral

Calculus Level pending

Define C ( α ) C(\alpha) to be the coefficient of x 1992 x^{1992} in the power series expansion about x = 0 x=0 of ( 1 + x ) α (1+x)^\alpha . Evaluate

∫ 0 1 C ( − y − 1 ) ( 1 y + 1 + 1 y + 2 + 1 y + 3 + ⋯ + 1 y + 1992 ) d y \large \int_{0}^1 C(-y-1) \left(\frac{1}{y+1}+\frac{1}{y+2}+\frac{1}{y+3}+\dots + \frac{1}{y+1992}\right)dy


The answer is 1992.

Vishruth Bharath by Vishruth Bharath , 17, USA

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1 solution

Tom Engelsman
Feb 8, 2018

Taken from the 1992 William J. Putnam Examination:

The usual Taylor series gives C ( − y − 1 ) = ( y + 1 ) ( y + 2 ) . . . . . ( y + 1992 ) / 1992 ! C(-y-1) = (y+1)(y+2).....(y+1992) / 1992! . Hence the integrand is simply the derivative of ( y + 1 ) ( y + 2 ) . . . ( y + 1992 ) / 1992 ! (y+1)(y+2) ... (y+1992)/1992! , so the integral evaluates to 1993 − 1 = 1992 . 1993 - 1 = \boxed{1992}.

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