Functions and things and integrals

Calculus Level pending

Define the functions { f n } n 0 \{ f_n\} _ {n \geq 0} as follows:

(1) f 0 ( x ) = Γ ( x + 1 x ) , and f_0(x) = \Gamma \left(\frac{x+1}{x}\right) \text{, and}

(2) f j ( x ) = x 2 d f j 1 ( x ) d x , for j > 0. f_j(x)=x^2 \frac{d f_{j-1}(x)}{d x} \text{, for } j>0.

lim x f 6 ( x ) = γ 4 π 2 a b + γ 2 π 4 c d + π 6 g h + j γ ( 2 γ 2 + π 2 ) ζ ( 3 ) + k ζ ( 3 ) 2 + m γ ζ ( 5 ) + γ 6 \underset{x\to \infty }{\text{lim}}f_6(x)=\frac{\gamma ^4 \pi ^2 a}{b}+\frac{\gamma ^2 \pi ^4 c}{d}+\frac{\pi ^6 g}{h}+j \gamma \left(2 \gamma ^2+\pi ^2\right) \zeta (3)+k \zeta (3)^2+m \gamma \zeta (5)+\gamma ^6

where a , b , c , d , g , h , j , k , m a,b,c,d,g,h,j,k,m are positive integers. Submit a + b + c + d + g + h + j + k + m a + b + c + d + g + h + j + k + m .

Bonus: Find closed form for

lim x f k ( x ) , k > 0 \underset{x\to \infty }{\text{lim}}f_k(x), k > 0

in terms of k , π , γ k, \pi, \gamma (the Euler–Mascheroni constant), and the zeta function ζ \zeta with positive odd integer arguments.


The answer is 453.

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1 solution

Hint: Imagine f 0 ( x ) = Γ ( x + 1 x ) f_0(x) = \Gamma(\frac{x+1}{x}) is expanded in a Laurent Series. Then lim x f n ( x ) \underset{x\to \infty}{\text{lim}}f_n(x) is the coefficient of 1 x n \frac{1}{x^n} (why?). There is a general formula for this coefficient involving only f 0 f_0 .

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