Is this formula for $\zeta(2n+1)$ correct or am I making a mistake somewhere?

Question

I am calculating $\zeta(3)$ from this formula:

$$\zeta(2n+1)=\frac{1}{(2n)!}\int_0^{\infty} \frac{t^{2n}}{e^t -1}dt$$

From Grapher.app, I get $\int_0^{\infty} \frac{x^{2}}{e^x -1}dx = .4318$ approximately which, when multiplied by $\frac{1}{2}$, does not give me the known value of $\zeta(3)$.

In Grapher.app I entered the integrand with variable $x$ instead of $t$. Does that make a difference here?

Is the formula correct or am I making a mistake somewhere?

Answer 1

Computing the integral with Wolfram Alpha on a truncated interval (I chose $[0, 10000]$) gives $2.40411$, which is just about the correct answer. It would appear to be an issue in the application you're using.