Codex @codex  0

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

The Beta function obeysThese identities follow immediately from the beta-gamma identity and the Gamma function recurrence.

On $(0,\infty )$, the Gamma function is the unique positive function $f$ satisfying $f(1)=1$, $f(x+1)=xf(x)$, and the condition that $\log f$ is a convex function. This last condition concerns the natural logarithm of $f$, rather than $f$ itself, and removes the interpolation freedom left by the Gamma function recurrence.^[ref]

The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

The beta-gamma identity gives $\Gamma (z/2{)}^2=B(z/2,z/2)\Gamma (z)$ for $\mathrm{Re}z>0$. A zero of $\Gamma (z)$ would therefore force zeros at every $z/2^m$. But $w\Gamma (w)=\Gamma (1+w)\to 1$ as $w\to 0$ in the right half-plane, a contradiction. The Gamma function recurrence then extends nonvanishing to every point where the meromorphic gamma function is finite. Nonpositive integers are poles, not zeros.

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

Consider $-y^{\prime \prime }-(n-1)y^{\prime }/x+x^{2}y=\lambda y$ for a positive integer $n$, with regular radial data at zero and square-integrable decay at infinity in the measure $x^{n-1}dx$. Its Sturm-Liouville form has $p=w=x^{n-1}$ and $q=x^{n+1}$. For the Gaussian trial $y=e^{a{x}^2}$, $a<0$, the Rayleigh quotient isThe Gamma function recurrence evaluates the ratio, and the minimum is $n$ at $a=-1/2$. It is the exact lowest eigenvalue: after the endpoint term $[x^n{y}^2{]}_0^{\infty }$ vanishes, integration by parts givesEquality gives the displayed positive Gaussian eigenfunction. For $n=1$, the radial regularity condition is the Neumann boundary condition at zero, not a vanishing value there. Changing the boundary condition changes the lowest eigenvalue and whether this trial is admissible.

Here $\psi ={\Gamma }^{\prime }/\Gamma$ is the digamma function, $B=\log 2+\frac{1}{2}\log \pi -1-\gamma /2$, and $\gamma$ is the Euler--Mascheroni constant. The sum runs over Nontrivial zeros of the Riemann zeta function with multiplicities. Each displayed paired summand is $O_s(|\rho {|}^{-2})$, so it converges locally uniformly away from zeros. Differentiate the genus-one Hadamard factorization of the Riemann xi function and use the Gamma function recurrence. Unlike an unpaired reciprocal-zero sum, this expression does not require an unstated summation convention.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.

Take $M^2>0$ initially to avoid an infrared divergence. A Schwinger parameterization and the Euclidean Gaussian integral give, in the convergent range $0<\mathrm{Re}d<4$,Integration over $t$ supplies the Gamma function. Its analytic continuation defines the answer in dimensional regularization:Use $d=4-ϵ$ near four dimensions and $d=6-ϵ$ near six dimensions. The Gamma function recurrence implies $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. ThereforeBoth are simple poles. Equivalently their residues as functions of $d$ are $-2/(4\pi {)}^2$ at $d=4$ and $2M^2/(4\pi {)}^3$ at $d=6$. The negative pole near six dimensions is a feature of analytic continuation; it is not a negative value of the original convergent positive integral. A massless zero-momentum integral is scaleless and cannot be used indiscriminately to extract these ultraviolet poles, because dimensional regularization then combines ultraviolet and infrared contributions.

For the following Feynman diagrams, specify the Euclidean action and additive counterterms explicitly so their signs are unambiguous. The four-dimensional interaction is ${\mu }^{ϵ}\lambda {\varphi }^4/4!$, the six-dimensional interaction is ${\mu }^{ϵ/2}g{\varphi }^3/3!$, and the common free inverse scalar propagator is $p^2+m^2$. Here $ϵ=4-d$ or $6-d$, respectively. The factors of $\mu$ give dimensionless renormalized couplings and change no leading pole residues. The cubic Euclidean theory is treated as a formal perturbation expansion about its massive quadratic scalar field theory; its real cubic potential is not bounded below.

Separate the constant embedding coordinate from its nonzero modes. Its integral gives the momentum-conservation delta function; below this is understood and absorbed into the amplitude normalization on the momentum-conserving locus. For the nonzero modes the Euclidean Polyakov action is quadratic, with worldsheet Green functionTreat the tachyon vertex operators as normal ordered, removing coincident self-contractions. The Gaussian functional integral, or the same calculation by Wick's theorem, givesAll remaining determinants are independent of the insertion points and momenta and enter the overall constant. This derives the Koba-Nielsen factor and the requested position integral.

For the Möbius transformation $f(z)=(az+b)/(cz+d)$, $ad-bc=1$, one hasThe transformed Koba-Nielsen factor contributes, at insertion $r$, the power $-{\alpha }^{\prime }{\sum }_{s\ne r}{k}_r\cdot k_s$ of $|c{z}_r+d|$. Four-momentum conservation makes this power ${\alpha }^{\prime }{k}_r^2$. The complete differential integrand therefore transforms byIt is invariant when ${\alpha }^{\prime }{k}_r^2=4$ for every external tachyon. The correlator alone is covariant; the integration measures are necessary for invariance. The group acting faithfully on the sphere is $PSL(2,\mathbb{C})$, with the two elements $\pm I$ of $SL(2,\mathbb{C})$ acting identically.

The apparent integral over all positions includes the residual conformal-group gauge volume. Divide by it and fix three distinct insertions. The Faddeev-Popov determinant, equivalently the three-point worldsheet ghost correlator and its conjugate, contributesSet $z_1=\Lambda$, $z_2=0$, $z_4=1$, then send $\Lambda$ to infinity on the sphere; the sign of the real direction is irrelevant. The factors involving $z_1$ behave as $|\Lambda {|}^{{\alpha }^{\prime }{k}_1\cdot {\sum }_{s\ne 1}{k}_s}=|\Lambda {|}^{-4}$, while $J\sim |\Lambda {|}^4$. They cancel, leaving $z_3=z$ as the one integration variable.

The remaining distance exponents areFor example $(k_2+k_3{)}^2=(k_1+k_4{)}^2=-t$ and each individual squared momentum is $4/{\alpha }^{\prime }$. AlsoConsequently the reduced amplitude is proportional toApply the supplied complex beta integral with $A=4+{\alpha }^{\prime }t/2$, $B=4+{\alpha }^{\prime }s/2$, and $C=4-A-B=4+{\alpha }^{\prime }u/2$. Its numerator arguments become $-1-{\alpha }^{\prime }t/4$, $-1-{\alpha }^{\prime }s/4$, $-1-{\alpha }^{\prime }u/4$, while its denominator arguments become $2+{\alpha }^{\prime }u/4$, $2+{\alpha }^{\prime }t/4$, $2+{\alpha }^{\prime }s/4$. Thus the four-point amplitude is the Virasoro–Shapiro amplitudewhere $C_0$ is momentum-independent. The integral converges initially when $\mathrm{Re}A<2$, $\mathrm{Re}B<2$, and $\mathrm{Re}(A+B)>2$; the displayed answer elsewhere is its analytic continuation.

The gamma function has simple poles at nonpositive integers and no zeros. At generic fixed $t$, the amplitude therefore has simple $s$-channel poles at $s=4(n-1)/{\alpha }^{\prime }$, $n=0,1,2,\dots$, with the same towers in $t$ and $u$ by crossing symmetry. They describe exchange of the closed string tachyon, massless states, and the infinitely many massive levels. The reciprocal denominator factors can give zeros at negative values $s=-4(\ell +2)/{\alpha }^{\prime }$; special kinematics can also make an exchange residue vanish.

To see the structure of a generic residue, set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$, so $a+b+c=1$. At $a=-n$, the Gamma function recurrence reduces the remaining factors to a constant times ${\prod }_{j=1}^n(b-j{)}^2$. The Virasoro–Shapiro amplitude pole residue is thus a polynomial of degree $2n$ in the other channel invariant, consistent with exchange up to spin $2n$. Crossed numerator poles must not simply be multiplied: near $a=-n+\delta$ and $b=-m+\epsilon$, $c=1+n+m-\delta -\epsilon$, and the zero of $1/\Gamma (1-c)$ cancels the putative double pole. The leading singular part is proportional toThe tree amplitude is meromorphic, with no loop threshold branch cuts; its channel singularities express factorization and string dual resonance.

Set $t=x^2$, so $dx=dt/(2\sqrt{t})$. ThenThe beta-gamma identity, the Gamma function recurrence and the gamma reflection formula give

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta-gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.

Write $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$ and ${\psi }_{\Gamma }={\Gamma }^{\prime }/\Gamma$. The global partial-fraction expansion of the zeta logarithmic derivative isHere $\gamma$ is the Euler--Mascheroni constant, and the sum includes multiplicities of the Nontrivial zeros of the Riemann zeta function. The paired terms are locally absolutely convergent away from their poles, by the Riemann–von Mangoldt formula. This identity follows by differentiating the genus-one Hadamard factorization of the Riemann xi function and using the Gamma function recurrence.

We first prove the classical zero-free region of the Riemann zeta function. The Euler-product nonvanishing of the Riemann zeta function and its functional equation place every nontrivial zero in the closed strip $0\le \beta \le 1$; no assertion about zeros on its boundary is needed yet. The real logarithmic derivative of Riemann xi and the digamma function asymptotic imply, for $1<\sigma \le 2$ and $|t|\ge 1$,Every summand being subtracted is nonnegative. If $\beta +it$ is a zero, retaining its summand gives $\Re F(\sigma +it)\le C\log (|t|+2)-1/(\sigma -\beta )$; at height $2t$ the upper bound $C\log (|t|+2)$ suffices. At real $\sigma$ the simple pole of $\zeta$ gives $F(\sigma )=1/(\sigma -1)+O(1)$.

For $\sigma >1$, the Dirichlet series $F(s)={\sum }_n\Lambda (n)n^{-s}$ has nonnegative coefficients. The identity $3+4\cos u+\cos 2u=2(1+\cos u{)}^2\ge 0$ therefore gives the three-four-one zero-free-region argumentWith $L=\log (|t|+2)$, a zero at $\beta +it$ would consequently satisfyChoose $\sigma =1+a/L$, with an absolute $a>0$ sufficiently small, and put $c=a/8$. If $\beta \ge 1-c/L$, this inequality would imply $4/(a+c)\le 3/a+C$. But $4/(a+a/8)-3/a=5/(9a)>C$ for small enough $a$, a contradiction.

For any fixed $t\ne 0$, the same positivity inequality as $\sigma \downarrow 1$ excludes a zero at $1+it$: its negative contribution has coefficient at least four, whereas the real pole contributes only three. Any zero at $1+2it$ only strengthens the contradiction. The pole at $s=1$ has a zero-free punctured neighbourhood. By compactness, the remaining bounded-height portion of the line has a zero-free neighbourhood as well. Shrink $c$ to combine these facts with the large-height bound. Hencewith the understanding that $s=1$ is a pole.

The Riemann–von Mangoldt explicit formula for the Second Chebyshev function, with half weight at a prime-power endpoint, isThe zero sum uses symmetric height truncation. To estimate it, take a half-integer $x$ and use the truncated explicit formula for the second Chebyshev functionThe zero-free region gives $|x^{\rho }|\le x\exp (-c\log x/\log (T+2))$ for these zeros. The Riemann–von Mangoldt formula and partial summation give ${\sum }_{|\Im \rho |\le T}|\rho {|}^{-1}\ll {\log }^2(T+2)$. Choose $T=\exp (\sqrt{\log x})$. Both the zero sum and the truncation error are then $O(x{e}^{-c^{\prime }\sqrt{\log x}})$ after reducing the positive constant to absorb logarithmic factors. Replacing an arbitrary $X$ by $\lfloor X\rfloor +1/2$ preserves the Von Mangoldt function sum and changes the main term by at most one. This proves the requested form of the Prime number theorem:

With $Z[J]=e^{iW[J]}$, the logarithm $W$ is the connected generating functional. Its derivative $\phi =\delta W/\delta J$ is the mean field. The specified Legendre transform satisfies $\delta \Gamma /\delta \phi =J$, and the quantum effective action generates amputated one-particle-irreducible vertices by further functional derivatives. In this question $\Gamma =-W+\int J\phi$ is the negative of the commonly used Minkowski effective-action convention; consequently its tree term is $-S[\phi ]$. Its diagrammatic content is unchanged if the convention is used consistently.

Expand the action around a stationary background in the presence of the source. For $\varphi =\phi +f$, integration by parts givesThe linear term vanishes at the tree-level saddle, $\delta S/\delta \phi +J=0$. Keeping the Gaussian fluctuation approximation, and fixing the field-independent normalization, givesto one-loop order. The shift from the tree saddle to the exact mean field cancels from the tree Legendre transform at this order. Hence the one-loop scalar effective action in the question's convention isThe functional determinant has a Feynman i-epsilon prescription before Wick rotation. Higher loops come from the omitted cubic and higher fluctuation vertices.

To see the one-loop one-particle-irreducible Feynman diagrams, write $\Delta ={\Delta }_0+U(\phi )$. Its determinant expansion isEach trace closes a single scalar loop, with propagators ${\Delta }_0^{-1}$ and background insertions $U$. No internal edge is a bridge, so these graphs are one-particle irreducible. Expanding the background factors produces their external legs.

For a constant $\phi$, set $A=V^{\prime \prime }(\phi )>0$. The quantum effective potential is defined here by $\Gamma [\phi ]=\int d^{d}x{V}_{\mathrm{eff}}(\phi )$. A plane-wave basis diagonalizes $\Delta$, and the supplied Wick rotation converts its correction per unit volume toHere the power of the renormalization scale keeps the continued integral in four-dimensional units. The Gamma function recurrence and expansion giveConsequentlyThe divergence is local, so subtract it with a potential counterterm. In the modified minimal subtraction scheme, remove its associated $\gamma -\log 4\pi$ constants as well. The one-loop scalar effective potential becomesup to an arbitrary field-independent vacuum-energy constant. Another allowed finite subtraction replaces the nonlogarithmic constant. Restoring a loop-counting parameter would multiply the displayed correction by $\hslash$. The logarithm assumes positive curvature $A$; a background with negative curvature requires an analytic continuation and can have an imaginary part signalling instability.

The quartic case is special because of quartic closure of scalar effective-potential counterterms. If $V$ has degree at most four, then $V^{\prime \prime }$ has degree at most two and $(V^{\prime \prime }{)}^2$ has degree at most four: its pole is removable by the original polynomial couplings and a constant. For example, $V=m^2{\phi }^2/2+\lambda {\phi }^4/4!$ givesThe last two terms reproduce the one-loop massive phi-fourth counterterms, and the first is a vacuum-energy subtraction. A degree-$r$ potential with $r>4$ instead generates degree $2r-4>r$, forcing new operators. This is the potential-level manifestation of power counting in quantum field theory and four-dimensional renormalizability, even though the finite effective potential itself need not remain polynomial.

At $x=0$, both denominators are nonzero because $a\notin 2\pi \mathbb{Z}$; they remain nonzero for every real $x\ge 0$. The integrand is therefore bounded on each finite interval. At infinity its terms decay respectively as $O(e^{-(1+b)x})$ and $O(e^{-(1-b)x})$, so $0<b<1$ gives absolute convergence.

Set $t=e^{-x}$. The two integrals becomeBy the Euler integral for the hypergeometric function and the Gamma function recurrence, ${\int }_0^1{t}^{s-1}/(1-zt)dt={}_2{F}_1(1,s;s+1;z)/s$ for $s>0$ and the indicated unit-circle values $z\ne 1$. HenceThe Gauss hypergeometric functions here use their continuation from the Euler integral, so no pole crosses the integration interval.

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.