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Question Number 59193 by salahahmed last updated on 05/May/19

	Answered by MJS last updated on 06/May/19
	$for x∈N ∫_0 ^∞ t^x e^(−t) dt=x! ⇒ we have one obvious solution at x=5 x^3 −x=x! (x−1)x(x+1)=x! ⇒ x≠0∧x≠1 (x−1)x(x+1)=x(x−1)(x−2)(x−3)... (x+1)=(x−2) ⇒ no solution remember we′re seeking x∈N ⇒ we only have to try factors of the constant (x+1)=(x−2)(x−3) x^2 ...+5=0 ⇒ x=5 (x+1)=(x−2)(x−3)(x−4) x^3 ...−25=0 ⇒ x=5 above x=5 x!>x^3 −x there are solutions for x∈R\Z approximating with a calculator I found x≈1.37439465 (±5×10^(−8) ) there are infinite solutions for x<−3 I found x.≈−3.02 and x≈−3.997 there should be solutions at x=−(2n−1)−δ_n and at x=−2n+ε with n>1∧n∈N and δ_n , ε_n being very small positive numbers$
	$for x \in N \underset{0}{\int^{\infty}} t^{x} e^{- t} d t = x!$ $\Rightarrow we have one obvious solution at x = 5$ $x^{3} - x = x!$ $(x - 1) x (x + 1) = x!$ $\Rightarrow x \neq 0 \land x \neq 1$ $(x - 1) x (x + 1) = x (x - 1) (x - 2) (x - 3) . . .$ $(x + 1) = (x - 2) \Rightarrow no solution$ $remember {we}^{'} re seeking x \in N \Rightarrow we only have$ $to try factors of the constant$ $(x + 1) = (x - 2) (x - 3)$ $x^{2} . . . + 5 = 0 \Rightarrow x = 5$ $(x + 1) = (x - 2) (x - 3) (x - 4)$ $x^{3} . . . - 25 = 0 \Rightarrow x = 5$ $above x = 5 x! > x^{3} - x$ $there are solutions for x \in R ∖ Z$ $approximating with a calculator I found$ $x \approx 1.37439465 (\pm 5 \times 10^{- 8})$ $there are infinite solutions for x < - 3$ $I found x . \approx - 3.02 and x \approx - 3.997$ $there should be solutions at x = - (2 n - 1) - δ_{n}$ $and at x = - 2 n + ϵ with n > 1 \land n \in N and$ $δ_{n}, ϵ_{n} being very small positive numbers$
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