For such an integral: $$\displaystyle \int_0^{\infty} \frac{t^m-t^n}{e^t-1} \, dt = 0$$ Given that m and n are constants and we suppose that m and n are equal, it can clearly be observed that the equation holds. The main issue would be if we ask ourselves if otherwise there's a unique counterexample of m and n such that m is not equal to n (especially for m and n in the critical strip) that can satisfy the equation. If the counterexample for Re(m) and Re(n) inside the critical strip that contradicts the property m=n doesn't exist, it can imply the Riemann hypothesis to be correct. Point to note: The integral was acquired from the Mellin transformation of the Riemann zeta function in the critical strip.
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- 1$\begingroup$ Since $$\mathcal{M}_t\left[\frac{1}{e^t-1}\right](s+1)=\int\limits_0^{\infty} \frac{t^s}{e^t-1} \, dt=\Gamma(s+1)\, \zeta(s+1)\,,\quad\Re(s)>0$$ by critical strip I assume you mean $-1<\Re(s)<0$ and are assuming analytic continuation of the integral? $\endgroup$Steven Clark– Steven Clark2025-06-02 01:41:37 +00:00Commented Jun 2 at 1:41
- 1$\begingroup$ If you plot $\Gamma(s+1)\, \zeta(s+1)$ in the interval $-1<s<0$, you'll see there are an infinite number of real values of $m$ and $n$ where $\Gamma(m+1)\, \zeta(m+1)=\Gamma(n+1)\, \zeta(n+1)$ and $-1<m<n<0$. Likewise if you plot $\Gamma(s+1)\, \zeta(s+1)$ for $s>0$, you'll see there are an infinite number of real values of $m$ and $n$ where $\Gamma(m+1)\, \zeta(m+1)=\Gamma(n+1)\, \zeta(n+1)$ and $0<m<n$. (see this WolframAlpha evaluation) $\endgroup$Steven Clark– Steven Clark2025-06-02 01:54:13 +00:00Commented Jun 2 at 1:54
- $\begingroup$ @StevenClark. First and foremost, you forgot to add the term 1/(1-2^1-s) for the Mellin transformation in the critical strip. Secondly, the correct critical strip bound is 0<Re(s)<1. Thirdly, kindly mention me any value of m and n in the critical strip where Re(m) isn't equal to Re(n) so that the integral in the question holds. $\endgroup$Kato Tyresse– Kato Tyresse2025-06-02 03:37:23 +00:00Commented Jun 2 at 3:37
- $\begingroup$ No, my formula is correct and consistent with your integral $$\int\limits_0^{\infty} \frac{t^m-t^n}{e^t-1} \, dt=\int\limits_0^{\infty} \frac{t^m}{e^t-1} \, dt-\int\limits_0^{\infty} \frac{t^n}{e^t-1} \, dt\\=\mathcal{M}_t\left[\frac{1}{e^t-1}\right](m+1)-\mathcal{M}_t\left[\frac{1}{e^t-1}\right](n+1)\\=\Gamma(m+1)\, \zeta(m+1)-\Gamma(n+1)\, \zeta(n+1)\,,\quad\Re(m)>0\land\Re(n)>0.$$ $\endgroup$Steven Clark– Steven Clark2025-06-02 04:38:03 +00:00Commented Jun 2 at 4:38
- $\begingroup$ The Mellin transform of $f(t)$ is defined as $$\mathcal{M}_t[f(t)](z)=\int\limits_0^{\infty} f(t)\, t^{z-1} \, dt$$ and you're evaluating it at $z=s+1$ (where $s$ can either be $m$ or $n$). $\endgroup$Steven Clark– Steven Clark2025-06-02 04:38:07 +00:00Commented Jun 2 at 4:38
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1 Answer
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The representation
$$\zeta(s)=\frac{1}{\Gamma(s)}\, \mathcal{M}_t\left[\frac{1}{e^t-1}\right](s)=\frac{1}{\Gamma(s)}\, \int\limits_0^{\infty} \frac{t^{s-1}}{e^t-1} \, dt\,,\quad\Re(s)>1\tag{1}$$ can be analytically extended to the critical strip $0<\Re(s)<1$ by formulas such as
$$\zeta(s)=\lim\limits_{N\to\infty} \left(\sum\limits_{n=1}^N \frac{1}{n^s}+\frac{1}{(s-1)\, N^{s-1}}\right)\tag{2}$$
and
$$\zeta(s)=\frac{1}{1-2^{1-s}}\, \lim\limits_{N\to\infty} \left(\sum\limits_{n=1}^N \frac{(-1)^{\,n-1}}{n^s}\right)\tag{3}.$$
The three intersections of the blue and orange curves in Figure (1) below correspond to three distinct values of $s$ in the critical strip with different real and imaginary parts where $\zeta(s)=i \frac{1}{2}$.
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Figure (1): Contour plots of $\Re(\zeta(s))=0$ and $\Im(\zeta(s))=\frac{1}{2}$ in blue and orange respectively
