Gamma Function Confusion

Calculus Level 3

$\begin{cases} f(x) = \displaystyle \int_0^{\infty} \left( t^{x+1} + t^{x} \right) e^{-t} \ dt \\ g(x)= \displaystyle \int_0^{\infty} \left( t^{x+1} - t^{x} \right) e^{-t} \ dt \\ h(x) = \dfrac{f(x) \times f(x-1) \times f(x-2) \times \cdots \times f(1)}{g(x) \times g(x-1) \times g(x-2) \times \cdots \times g(1)}\end{cases}$

Define $f(x)$ , $g(x)$ and $h(x)$ as above, then $2h(x)$ can be expressed as a polynomial for integral $x$ , which satisfy $x > 2$ . Enter the sum of the coefficients of this polynomial (including the constant term) to 2 decimal places.

Hint: Try to rephrase this in terms of the gamma function.

If you do not know how to solve this view this wiki .

The answer is 6.00.

1 solution

Chew-Seong Cheong
Nov 30, 2017

Relevant wiki: Gamma function & Factorial

Note that gamma function is defined as $\Gamma (s) = \displaystyle \int_0^\infty t^{s-1}e^{-t} dt$ . Therefore, we have:

$\begin{aligned} f(x) & = \int_0^\infty \left(t^{x+1}+t^x \right) e^{-t} dt = \Gamma (x+2) + \Gamma (x+1) = (x+1)\Gamma(x+1) + \Gamma(x+1) = (x+2)\Gamma(x+1) \\ g(x) & = \int_0^\infty \left(t^{x+1}-t^x \right) e^{-t} dt = \Gamma (x+2) - \Gamma (x+1) = (x+1)\Gamma(x+1) - \Gamma(x+1) = x\Gamma(x+1) \\ \implies \frac {f(x)}{g(x)} & = \frac {(x+2)\Gamma(x+1)}{x\Gamma(x+1)} = \frac {x+2}x \\ h(x) & = \frac {f(x)f(x-1)f(x-2)\cdots f(1)}{g(x)g(x-1)g(x-2)\cdots g(1)} \\ & = \frac {(x+2)(x+1)(x)\cdots (3)}{x(x-1)(x-2)\cdots (1)} \\ & = \frac {(x+2)!}{2x!} \\ \implies 2h(x) & = \frac {(x+2)!}{x!} = (x+2)(x+1) = x^2 + 3x + 2 \end{aligned}$

Therefore, the sum of coefficients of the polynomial is $1+3+2 = \boxed{6.00}$

Gamma Function Confusion

The answer is 6.00.

1 solution

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