$\prod_{i=1}^\infty\left(\frac{i+x}{i+1}\right)^{1/i}\stackrel{?}{=}x$

Question

$$\prod_{i=1}^\infty\left(\frac{i+x}{i+1}\right)^{1/i}\stackrel{?}{=}x$$

I do not have the knowledge needed to prove this (assuming it is true).

quick equivalent forms:

$$\sum_{i=1}^{\infty}\frac{\log(i+x)-\log(i+1)}{i}\stackrel{?}{=}\log(x) $$

$$\sum_{i=1}^{\infty}\frac{\log\left(1-\frac{1-x}{i+1}\right)}{i}\stackrel{?}{=}\log(x) $$

all the solutions I know of have $i$ in the base or the exponent, but not both. It almost looks like Evaluation of $\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}$, except the solutions depend on integer exponents and I have fractional exponents.

Answer 1

I don't think the equality generally holds, specially by differentiating both sides we must have $$\sum_{i=1}^{\infty} {1\over (i+x)\cdot i}={1\over x}$$which for $x\in\Bbb N-\{1\}$ means that $$\sum_{i=1}^{x}{1\over i}={1\over x}$$which doesn't hold.

Answer 2

The correct result should be

$$ \prod_{i=1}^\infty \left( \frac{i+x}{i+1} \right)^{1/i} = \exp \left(\int_1^x \frac{\Psi(t+1)+\gamma}{t} \; dt\right) $$

See my comment to Mostafa Ayaz's answer.

I don't know if this can be written in a more "closed-form" way than this, but it's certainly not $x$.