We know that, for example, the Gamma function is a perfect integral representation for the factorial $n!$ for a natural number $n$.

$$\Gamma[n] = \int_0^{+\infty} t^{n-1}e^{-t}\text{d}t = (n-1)!$$

Is there a similar integral representation through which I might find Fibonacci's numbers? Something like

$$F_n = \int_0^{+\infty} F(x, n)\ \text{d}x$$

to obtain the $n-$th Fibonacci's number?

P.s. Not necessarily an integration from zero to infinity.


One example can be found on the Wolfram functions site $$F_{2n}=\frac n2 \left(\frac32\right)^{n-1}\int_0^{\pi} \left(1+\frac{\sqrt 5}{3}\cos x\right)^{n-1} \sin x \,dx,$$ and another one in this note: $$F_n=\frac1{\sqrt5}\left(\frac{\sqrt5 +1}{2}\right)^n-\frac2\pi \int_0^{\infty}\frac{\sin\frac{x}2}{x}\frac{\cos n x-2\sin x\sin nx}{5\sin^2 x+\cos^2 x}dx.$$

  • $\begingroup$ This is what I was looking for, an integral way to find them. Thank you, that will surely help me! I will looking forward for other examples! $\endgroup$
    – Turing
    Dec 1 '15 at 12:54

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