Can i write $H_{n}=\frac{1}{n!} \int_{0}^{\infty} e^{-t} t^{n} \ln(t) dt$

Question

Consider this : $x!=x×(x-1)×(x-2)...2×1$ Differentiating w.r.t. $x$

$$\frac{d}{dx} (x!)=\bigg( (x-1)(x-2)..2.1)+(x(x-2)(x-3)...2.1)+...(x(x-1)(x-1)...2.1)\bigg)=\frac{1}{x!}\bigg(1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{x}\bigg)$$

And using the defination of gamma function LHS could be found out.

i.e.

$$H_{x}=\frac{1}{x!} \int_{0}^{\infty} e^{-t} t^{x} \ln(t) dt$$

Whats wrong in this approach ?

Answer 1

That's the correct formula for the derivative of the Gamma function yes. You can determine that by differentiation under the integral sign. Not sure what the point in dividing by $x!$ (whose meaning should be clarified in terms of $\Gamma$) then multiplying by it again is.

I'm not sure what that product means for real $x$. It's not equivalent to the Gamma function. If it's a product of $x$ integers (which I assume it is since it counts down from $x$ to $1$) differentiation with respect to $x$ does not make sense. Otherwise you're hiding too much behind the $\ldots$ and you should clarify. Note that if you look at an infinite product representation for $\Gamma$ for positive integers (for example the Euler form) you'll realise they simplify to the familiar finite factorial product, but are not as nice otherwise, so it's faulty logic to substitute non-integers in there.

Edit: Just realised you're talking about the harmonic numbers. No idea how you got the connection from that product. It does seem to be the case that:

$$\frac 1 {n!} \int_0^\infty e^{-t} t^n \ln t \mathrm d t + \gamma = H_n$$

Proof: We have:

$$\psi(z+1)=\frac {\Gamma'(z + 1)} {\Gamma(z+1)} = \frac {\left(\Gamma(z + 1)\right)'} {\Gamma(z + 1)} = \frac {\left(z \Gamma(z)\right)'} {z \Gamma(z)} = \frac {\Gamma(z) + z\Gamma'(z)} {z \Gamma(z)} = \frac 1 z + \psi(z)$$

So for integer $N$:

$$\psi(N+1) - \psi(N) = \frac 1 N$$

Summing from $N = 1$ to $N = n$ you obtain:

$$\psi(n + 1) - \psi(1) = \psi(n + 1) + \gamma = H_n$$

Note now that:

$$\psi(n + 1) = \frac {\Gamma'(n + 1)} {\Gamma(n + 1)} = \frac {\Gamma'(n + 1)} {n!}$$

Writing $\Gamma'(n + 1)$ as an integral via differentiation under the integral sign with the integral definition of $\Gamma$ (as mentioned earlier), gives the desired result.