Euler-Gauss limit and Weierstrass product for the gamma function

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According to Euler's definition, the gamma function Γ(x)\Gamma (x) is determined by the following formula:

Γ(x)=lim⁡n→∞n! nxx(x+1)⋯(x+n). \Gamma (x)=\lim _{n \rightarrow \infty } \frac{n! \, n^{x}}{x(x+1)\cdots (x+n)}.

Starting from this formula:

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(a)

Express the function Γ(x)\Gamma (x) in the form of an infinite product;

(b)

Prove that Γ(x)\Gamma (x) is meaningful for all real xx that are not negative integers;

(c)

Deduce the following property

Γ(x+1)=xΓ(x); \Gamma (x+1)=x\Gamma (x);
(d)

Find the value of Γ(n)\Gamma (n) for positive integer nn.

Prove the identity:

∏n=1∞n(a+b+n)(a+n)(b+n)=Γ(a+1)Γ(b+1)Γ(a+b+1). \prod _{n=1}^{\infty } \frac{n(a+b+n)}{(a+n)(b+n)} = \frac{\Gamma (a+1)\Gamma (b+1)}{\Gamma (a+b+1)}.
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Determine for which values of xx the infinite products converge. ∏n=1∞(1+xn)e−x/n\prod_{n = 1}^{\infty }\left(1+\frac{x}{n}\right) e^{-x / n}.

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Prove that:

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(1)

the infinite product Γ(x)=1x∏n=1∞(1+1/n)x1+x/n\Gamma (x) = \frac{1}{x} \prod_{n = 1}^{\infty } \frac{(1+1 / n)^{x}}{1+x / n} absolutely converges for all x≠m,m=0,−1,−2,…x \neq m, m = 0,-1,-2, \ldots;

(2)

Γ(x)=lim⁡n→∞n!nxx(x+1)(x+2)…(x+n)\Gamma (x) = \lim_{n \rightarrow \infty } \frac{n ! n^{x}}{x(x+1)(x+2) \ldots (x+n)};

(3)

Γ(x+1)=xΓ(x)\Gamma (x+1) = x \Gamma (x) (Euler).

Prove that 1Γ(x+1)=eCx∏n=1∞(1+xn)e−x/n\frac{1}{\Gamma (x+1)} = e^{C x} \prod_{n = 1}^{\infty }\left(1+\frac{x}{n}\right) e^{-x / n}, where CC is Euler's constant (Weierstrass).

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Prove that

∏n⩾0(1−a2(n+b)2)=Γ2(b)Γ(b+a)Γ(b−a) \prod _{n \geqslant 0}\left(1-\frac{a^{2}}{(n+b)^{2}}\right)=\frac{\Gamma ^{2}(b)}{\Gamma (b+a) \Gamma (b-a)}

(a,b∈R,b≠0,b±a≠0,−1,−2,−3,…)(a, b \in \mathbb {R}, b \neq 0, b \pm a \neq 0,-1,-2,-3, \ldots ). Deduce from this that the fraction Γ(x+12)xΓ(x)\frac{\Gamma \left(x+\frac{1}{2}\right)}{\sqrt{x} \Gamma (x)} is increasing on the interval (0, +∞+\infty).

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The Γ\Gamma-function is defined for α>0\alpha > 0 (or ℜα>0\Re \alpha > 0) by the integral

Γ(α)=∫0∞e−xxα−1 dx. \Gamma (\alpha ) = \int _{0}^{\infty } e^{-x} x^{\alpha -1} \, dx .

Using Problem 89 of Part I prove that

Γ(α)=lim⁡n→∞nα−1n!α(α+1)⋯(α+n−1),α>0. \Gamma (\alpha ) = \lim _{n \to \infty } \frac{n^{\alpha -1} n!}{\alpha (\alpha +1) \cdots (\alpha +n-1)}, \qquad \alpha > 0 .
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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.
[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.