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Removing the completion's poles produces an entire function with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Its zeros are the nontrivial zeros of the Riemann zeta function. Its finite order allows Hadamard factorization, while the xi modulus criterion for the Riemann hypothesis characterizes their horizontal location.

The Stefan–Boltzmann law gives the total hemispherical radiant exitance of an ideal blackbody at absolute temperature $T$:Here $\sigma$ is the Stefan-Boltzmann constant. Total means integrated over all wavelengths; hemispherical means integrated over all outward directions with the projected area weighting. Thus $M_b$ is emitted power per actual emitting area, rather than directional radiance or spectral radiance. Absolute temperature must be in kelvin; raising a Celsius reading to the fourth power gives the wrong result. A uniform surface of area $A$ emits radiant flux $P_{\mathrm{emit}}=A{M}_b$. A real surface instead has $M=ϵ(T)\sigma T^4$, where $ϵ(T)$ is its total hemispherical emissivity. The older symbol $F$ in the formula above denotes this emitted radiant exitance.^[ref]

For an ideal blackbody, doubling absolute temperature multiplies radiant exitance by $16$. With the approximate Stefan-Boltzmann constant $5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$, the radiant exitances at $300\mathrm{K}$ and $600\mathrm{K}$ are approximately $459.300$ and $7348.805\mathrm{W}{\mathrm{m}}^{-2}$. This steep response explains why thermal radiation becomes so consequential at high temperature. It also enables heat transfer through a vacuum, where thermal conduction through a gas and convection are absent.

A short derivation integrates Planck's law in its frequency form. For a blackbody, direction-independent spectral radiance is $B_{\nu }(T)=2h{\nu }^3/[c^2(e^{h\nu /(k_{B}T)}-1)]$, where $h$ is the Planck constant, $k_B$ the Boltzmann constant and $c$ the speed of light. The outward hemisphere contributes $\int \cos \theta d\Omega =\pi$, soExpanding $(e^x-1{)}^{-1}={\sum }_{n=1}^{\infty }{e}^{-nx}$, nonnegative terms allow termwise integration. Repeated integration by parts gives ${\int }_0^{\infty }{x}^3{e}^{-nx}dx=6/n^4$; the Riemann zeta function value $\zeta (4)={\pi }^4/90$ gives the remaining integral as ${\pi }^4/15$. Hence $M_b=\sigma T^4$ with the coefficient in Stefan-Boltzmann constant. The angular integral is essential: omitting it confuses radiance with radiant exitance.

For $\sigma >1$, the logarithm of the displayed product is a prime-power sum with coefficients $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. A zero of order $a\ge 1$ at $1+it$, with $t\ne 0$, would make the product vanish like $O((\sigma -1{)}^{4a-3})$: the real factor has a pole of order three, and the third factor is bounded. This contradicts its lower bound one. Hence the Riemann zeta function has no zeros on its line of real part one; its pole at one is not a zero.

The totient divisor-sum identity and multiplication of absolutely convergent Dirichlet series give the quotient for $\mathrm{Re}s>2$. The bound $\varphi (n)\le n$ proves convergence there. The simple pole of the Riemann zeta function at one makes the quotient unbounded as real $s\downarrow 2$. Since the coefficients are nonnegative, convergence at two would bound all these larger-real-part values, a contradiction. Thus the abscissa of absolute convergence of a Dirichlet series is two.

For sufficiently large $|t|$, the Riemann zeta function has no zeros in the displayed region, for a fixed positive $c$. To derive it from the Richert bound for the Riemann zeta function, take $\eta \asymp (\log \log t/\log t{)}^{2/3}$. The logarithm of the maximum on the two discs in the Landau zero-free-region theorem is $O(\log \log t)$, as is $\log (1/\eta )$. Dividing $\eta$ by this logarithmic factor gives the displayed width.

Let $X$ have survival function $\mathbb{P}(X\ge x)=x^{-s}$ for $x\ge 1$, with $s>1$. Its expected value is $s/(s-1)$. The tail-sum formula for expectation givesThe absolutely continuous probability distribution gives zero mass to the integers, so $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely and $\mathbb{E}[\lceil X\rceil ]=\zeta (s)+1$. Taking expectations of $\lfloor X\rfloor \le X\le \lceil X\rceil$ gives $\zeta (s)\le s/(s-1)\le \zeta (s)+1$, which rearranges to the displayed bounds on the Riemann zeta function. Finiteness is justified by $\lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$.

For a real exponent $s>1$ and a positive integer $N$, the series truncation error in the defining Riemann zeta function series isThe integral comparison for the positive decreasing function $x^{-s}$ gives the explicit boundsOn each unit interval, the area under $x^{-s}$ lies between its endpoint heights. Summing from the first omitted term gives the lower integral from $N+1$ to infinity and the upper integral from $N$ to infinity. This uses positivity and monotonicity, not an assumption that the last computed term equals the remaining error.

At $N=10$, the bounds are $1/11\le R_{10}(2)\le 1/10$, whereas $1/(3\cdot 11^3)\le R_{10}(4)\le 1/(3\cdot 10^3)$. The errors above fit both intervals. Merely looking at the eleventh term would give $1/121\approx 0.008264$ for squares, far smaller than the actual remaining $0.095166$.

For a chosen absolute error target $\epsilon >0$, the sufficient condition is $N\ge ((s-1)\epsilon {)}^{-1/(s-1)}$, rounded upward to an integer. Thus $\epsilon =10^{-4}$ requires $N=10000$ by this bound for squares but only $N=15$ for fourth powers. This is a guarantee from an upper bound, not a claim that those are the smallest successful term counts.