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Substitution ofand integration by parts giveThus takeand require the endpoint term to vanish. Infinite contour ends must lie in sectors where $\mathrm{Re}(t^4)>0$, namely sectors centred on the positive and negative real and imaginary axes, each of angular width $\pi /4$.

Let the common final segment be the positive real ray from $0$ to $+\infty$, and choose the initial rays from $-\infty$, $+i\infty$, and $-i\infty$ to $0$. Call the resulting contours ${\gamma }_1,{\gamma }_2,{\gamma }_3$. PutSince $w_i^{(r)}(0)={\int }_{{\gamma }_i}{t}^r{e}^{-t^4/4}dt$, for $r=0,1,2$ the three rows areConsequentlyUsing the Gamma function reflection identity,The initial-data vectors are therefore linearly independent, so the three contour integrals are linearly independent solutions.

Taking reciprocals in the finite product defining the gamma function givesLet $H_n={\sum }_{k=1}^n{k}^{-1}$. Sinceand $H_n-\log n\to \gamma$, where $\gamma$ is the Euler--Mascheroni constant, passage to the limit gives the Weierstrass product for the reciprocal gamma function

The logarithmic derivative of this identity isSince $k^{-1}-(z+k{)}^{-1}=z/[k(z+k)]$, the digamma function therefore satisfies

For real $z>0$, termwise differentiation gives the trigamma functionThus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping givesThe intermediate value theorem supplies a zero in $(1,2)$, and strict increase makes it unique. This is the positive zero of the digamma function.

Although $x\mapsto |x{|}^{-1}$ is not integrable on all of ${\mathbb{R}}^3$, it defines a tempered distribution. For $x\ne 0$, the Gamma function integral givesThe three-dimensional form of the Fourier transform of a Gaussian isTherefore, up to a nonzero constant depending on the Fourier-transform convention,For $\xi \ne 0$, the substitution $u=|\xi {|}^2/(4t)$ turns the last integral intoThus, as asserted by the Fourier transform of inverse distance in three dimensions,as a tempered distribution, where $C^{\prime }\ne 0$.

For $\Re s>1$, absolute convergence permits separation into odd and even terms:Using the defining integral of the Gamma function and the substitution $u=nt$ givesTermwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum isConsequently the Dirichlet eta function satisfies

Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thusprovides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use insteadwith an integer $k$ for which $1-k^{1-s}\ne 0$; the bounded partial sums of $a_n$ give convergence for $\Re s>0$. This shows that those apparent singularities are removable and yields the Analytic continuation of the Riemann zeta function to the right half-plane.

At $s=1$,whereasHenceso $s=1$ is a simple pole and its residue is

For $t>0$, the product from part (b) has $q=e^{-2\pi t}\in (0,1)$, soThe weight-twelve transformation law givesTogether with exponential decay as $t\to \infty$, this implies rapid decay at both endpoints for the Mellin transformThe integral therefore converges for every real $s$, and its integrand is strictly positive.

In the half-plane where the Dirichlet series may be integrated term by term,Analytic continuation preserves this identity. For real $s>0$, both $(2\pi {)}^s$ and the Gamma function $\Gamma (s)$ are positive, soThis is the positivity statement in the Mellin transform of the modular discriminant, and in particular $L(\Delta ,s)\ne 0$.

Because the gamma function has simple poles at the nonpositive integers and no zeros, the $s$-dependent numerator $\Gamma (-{\alpha }^{\prime }s/4)$ has poles atorFactorization of a scattering amplitude identifies each pole with an intermediate on-shell state. The Type II superstring mass spectrum therefore contains a massless level at $n=0$ and an infinite equally spaced tower in squared mass for $n\ge 1$. There is no negative-$M^2$ pole, consistently with the absence of a tachyon in Type II superstring theory.

A galactic distribution function $f(\mathbf{x},\mathbf{v},t)$ is the stellar mass or number per six-dimensional phase space volume,Its velocity moments give the spatial density, mean velocity, and velocity dispersion; integrating those quantities along the line of sight and weighting by luminosity produces surface-brightness and line-of-sight-velocity observables. A model is compared with data only after the same projection, selection function, and instrumental convolution have been applied.

Jeans theorem states that every steady solution of the Collisionless Boltzmann equation depends on phase-space coordinates only through integrals of motion. Conversely, every nonnegative function of isolating integrals is a steady collisionless distribution function on the region where those integrals are defined.

For the stated power law in relative energy, isotropy givesWith the requested change of variables $v=\sqrt{2\Psi }\cos \theta$, the density becomeswhere the last equality uses the Beta function and Gamma function. Thusfor $n>1/2$.

The normalized second velocity moment isThe same substitution and the Beta-function recurrence giveConsequently the one-dimensional isotropic velocity dispersion is ${\sigma }^2=\overline{v^2}/3=\Psi /(n+1)$, proving the required linear dependence on the relative potential.

The meromorphic continuations of $F$ and the gamma function have the same recurrence and the same value at $1$. Hence their residues at every nonpositive integer are the same. Their difference $f=F-\Gamma$ therefore has removable singularities there and is analytic elsewhere, so it extends to an entire function. Equivalently, $f(z+1)=zf(z)$ and $f(1)=0$ show inductively that the apparent quotient singularities are removable.

The probability density function must integrate to one, soAlso, linearity of expectation and the mean and variance of the standard normal distribution giveUsing the probability density function to compute this expected value givesThese are the values $\Gamma (1/2)$ and $\Gamma (3/2)$ of the gamma function.

Put $u=x^2$, so $xdx=du/2$. Then integration by substitution givesThe last two integrals are elementary exponential moments, or values of the Gamma function.

In the Gamma function integral at $p=1/2$, use $x=t^2$, $dx=2tdt$, with $t>0$. Then $x^{-1/2}dx=2dt$, soThe substitution can first be made on finite positive intervals and then extended by monotone limits, justifying the endpoint passage.

For $n>1$, integration by parts with $u=x^{n-1}$ and $dv=e^{-x}dx$ yieldsThe boundary term vanishes at zero because $n-1>0$, and at infinity because exponential decay dominates every fixed power. The initial improper integral is $I_1={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating the recurrence relation therefore givesThis is the positive-integer case of the Gamma function identity $\Gamma (n)=(n-1)!$.

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly:

For $\mathrm{Re}z>0$, integration by parts givesThe boundary term vanishes because $t^{\mathrm{Re}z}\to 0$ at zero and exponential decay dominates any fixed power at infinity. ThusThis also gives $\Gamma (n+1)=n!$ for nonnegative integers, linking the Gamma function to the factorial.

Substituting $z=1$ into the defining gamma integral givesBoth endpoints are convergent, so no continuation of the Gamma function is needed here.

Put $x=e^{-t}$. Thenwhere the second substitution is $u=(k+1)t$ and the final integral is the given gamma function value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the exponential function expansion givesEvery term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and henceThis identity is one form of Sophomore's dream.

For $\mathrm{Im}k>0$, put $s=-ik$, so $\mathrm{Re}s>0$. The Laplace transform for the Gamma function, analytically continued from positive real $s$, givesChoose $\arg k\in (0,\pi )$ in the upper half-plane. Since $\arg (-ik)=\arg k-\pi /2$,This integral is absolutely convergent throughout the upper half-plane. On the nonzero real axis it converges conditionally by Dirichlet test for improper integrals, and its value is the upper boundary limit of the same branch. In particular negative real $k$ has argument $\pi$, not $-\pi$ in this formula.

At $k=0$ the integral diverges at infinity; in the lower half-plane its exponential function factor grows and the ordinary improper integral diverges. The origin itself causes no local-integrability problem for $x^{-1/4}$. A useful continuation of the chosen power uses a branch cut descending from $k=0$ with argument in $(-\pi /2,3\pi /2)$.

On the selected steepest descent contour put $t=e^{i\pi /6}r$, $r\ge 0$. Then $i{t}^3=-r^3$ and $dt=e^{i\pi /6}dr$. Its exact contribution isUse the Gamma function integral and its recurrence $\Gamma (4/3)=\Gamma (1/3)/3$ to obtainThere is no remainder here: the complete infinite descent ray was integrated exactly.

Integrating $e^{tx}$ against the Gamma density gives the moment-generating function of a gamma distribution:The change of variables $y=(\lambda -t)x$ reduces the integral to $\Gamma (r)=(r-1)!$ using the gamma function. For $t\ge \lambda$ the defining integral is infinite. In particular the moment-generating function is finite on a neighbourhood of zero.

Assume $p<1/2$, and put $a=\theta /p$ and $r=p/q\in (0,1)$. The detailed balance equations giveMultiplying from $0$ to $k-1$, the last factor telescopes, while the product involving $a$ is evaluated by the Gamma function:Consequently the requested invariant distribution can be expressed using the paper's power series asThe assumed convergence of $g_{p,\theta }(r)$ ensures that this normalization is finite and positive.

For an explicit normalizing constant, the binomial series givesThe termwise differentiation of a power series is valid for $|r|<1$, soThe last equality uses $r=p/q$ and $a=\theta /p$: $ar+\theta (1-r)=\theta /q+\theta (q-p)/q=2\theta$. ThereforeThese positive terms sum to one and satisfy the detailed balance equations, so part (a)(ii) verifies stationarity. Irreducibility makes the invariant distribution unique.