Monday Math 29
Consider the function )
, 
defined by the integral: =\int_{0}^{\infty}\frac{t^{x-1}}{e^t-1}\,dt)
. Can we find a formula for in terms of known functions?
First, we note that for 
, we have the series 
. Thus 
. Making the change of variable 
, we find that for 
, 
. Now, setting 
, we see 
, for 
.
Thus
&=&\int_{0}^{\infty}\frac{t^{x-1}}{e^t-1}\,dt\\&=&\int_{0}^{\infty}t^{x-1}\left(\sum_{n=1}^{\infty}e^{-nt}\right)\,dt\\&=&\sum_{n=1}^{\infty}\left(\int_{0}^{\infty}t^{x-1}e^{-nt}\,dt\right)\end{eqnarray})
.
Now, for the integral in each term, 
, we make the substitution 
. Then 
and 
, so our integral becomes:
,
and you should recognize that last integral as the gamma function of x.
Therefore, we find:
&=&\sum_{n=1}^{\infty}\left(\int_{0}^{\infty}t^{x-1}e^{-nt}\,dt\right)\\&=&\sum_{n=1}^{\infty}\left(\frac{1}{n^x}\operatorname{\Gamma}(x)\right)\\&=&\operatorname{\Gamma}(x)\sum_{n=1}^{\infty}\frac{1}{n^x}\end{eqnarray})
,
and that last sum is also familiar: it is the Riemann zeta function.
Thus, we see that =\operatorname{\Gamma}(x)\operatorname{\zeta}(x))
Tags: Math, Monday Math, Gamma Function, Riemann Zeta Function
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