396 words
2 minutes
Ramanujan Master Theorem
2025-10-23
Mathematic
Mathematic
/
Complex
Prerequisites

This article requires knowledge of complex numbers and some transforms.

  • If you don’t know complex numbers, you can refer to this article.
  • If you don’t know transforms, you can refer to this article.

Shift Operator#

We define: D≡ddxD\equiv \frac{\mathrm d}{\mathrm dx}. Consider

eD=∑n=0∞Dnn!e^D=\sum_{n=0}^\infty\frac{D^n}{n!}

Acting on a function f(x)f(x), we have:

eDf=∑n=0∞Dnn!f(x)=∑n=0∞f(n)(x)n!1ne^Df=\sum_{n=0}^\infty\frac{D^n}{n!}f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x)}{n!}1^n

Noting that this is the Taylor expansion of the function f(x)f(x) at xx, we clearly have:

eDf=f(x+1)\boxed{e^Df=f(x+1)}

Similarly:

etDf=∑n=0∞f(n)(x)n!tn=f(x+t)e^{tD}f=\sum_{n=0}^\infty\frac{f^{(n)}(x)}{n!}t^n=f(x+t)

Consider computing:

eaD⋅ebD=∑n=0∞anDnn!⋅∑n=0∞bnDnn!=∑n=0∞Dn∑k=0nakbn−kk!(n−k)!=∑n=0∞Dnn!∑k=0n(nk)akbn−k⏟(a+b)n=∑n=0∞Dnn!(a+b)n=e(a+b)D\begin{aligned} e^{aD}\cdot e^{bD}&=\sum_{n=0}^\infty\frac{a^nD^n}{n!}\cdot\sum_{n=0}^\infty\frac{b^nD^n}{n!}\\ &=\sum_{n=0}^\infty D^n\sum_{k=0}^n\frac{a^k b^{n-k}}{k!(n-k)!}\\ &=\sum_{n=0}^\infty\frac{D^n}{n!}\underbrace{\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}}_{(a+b)^n}\\ &=\sum_{n=0}^\infty\frac{D^n}{n!}(a+b)^n\\ &=e^{(a+b)D} \end{aligned}

Thus the shift operator satisfies exponential operations.

Difference and Shift#

Define the difference operator Δ\Delta as:

Δf(x)=f(x+1)−f(x)=eDf(x)−f(x)=(eD−1)f(x)\Delta f(x)=f(x+1)-f(x)=e^Df(x)-f(x)=(e^D-1)f(x)

Thus:

Δ=eD−1\Delta=e^D-1

Taking the inverse on both sides, we have:

Δ−1=1eD−1=1DDeD−1\Delta^{-1}=\frac1{e^D-1}=\frac1D\frac D{e^D-1}

The latter part is exactly the generating function of Bernoulli numbers, so expanding gives:

Δ−1=1D∑n=0∞Bnn!Dn\Delta^{-1}=\frac1D\sum_{n=0}^\infty\frac{B_n}{n!}D^n

Note that:

∑=Δ−1,∫=D−1\sum=\Delta^{-1},\int=D^{-1}

Substituting into the above equation, we get:

∑=∫∑n=0∞Bnn!Dn=∫+∑n=1∞Bnn!∫Dn=∫+∑n=1∞Bnn!Dn−1\begin{aligned} \sum&=\int\sum_{n=0}^\infty\frac{B_n}{n!}D^n\\ &=\int+\sum_{n=1}^\infty\frac{B_n}{n!}\int D^n\\ &=\int+\sum_{n=1}^\infty\frac{B_n}{n!}D^{n-1}\\ \end{aligned}

Noting that the odd-indexed Bernoulli numbers (except the first term) are all 00, we have:

∑=∫−12+∑n=1∞B2n(2n)!D2n−1\sum=\int-\frac12+\sum_{n=1}^\infty\frac{B_{2n}}{(2n)!}D^{2n-1}

Applying this formula from aa to bb to the function f(x)f(x), we obtain:

∑a≤n≤bf(n)=∫abf(x)dx−f(b)−f(a)2+∑n=1∞B2n(2n)![f(2n−1)(b)−f(2n−1)(a)]\sum_{a\le n\le b}f(n)=\int_a^bf(x)\mathrm dx-\frac{f(b)-f(a)}2+\sum_{n=1}^\infty\frac{B_{2n}}{(2n)!}\left[f^{(2n-1)}(b)-f^{(2n-1)}(a)\right]

Thus simply deriving the Euler-Maclaurin summation formula.

Ramanujan Master Theorem#

Consider the Laplace transform:

L{xs−1}(eD)=∫0∞xs−1e−xeD dx=Γ(s)e−sD\mathcal L\{x^{s-1}\}(e^D)=\int_0^\infty x^{s-1}e^{-xe^D}\,\mathrm dx=\Gamma(s)e^{-sD}

We substitute

e−xeD=∑n=0∞enDn!(−x)ne^{-xe^D}=\sum_{n=0}^\infty\frac{e^{nD}}{n!}(-x)^n

and let it act on the function ϕ(0)\phi(0), yielding:

∫0∞xs−1∑n=0∞enDn!(−x)nϕ(0) dx=Γ(s)e−sD⋅ϕ(0)⏟ϕ(−s)\int_0^\infty x^{s-1}\sum_{n=0}^\infty\frac{e^{nD}}{n!}(-x)^n\phi(0)\,\mathrm dx=\Gamma(s)\underbrace{e^{-sD}\cdot\phi(0)}_{\phi(-s)}

Let

f(x)=∑n=0∞enDϕ(0)n!(−x)n=∑n=0∞ϕ(n)n!(−x)n\begin{aligned} f(x)&=\sum_{n=0}^\infty\frac{e^{nD}\phi(0)}{n!}(-x)^n\\ &=\sum_{n=0}^\infty\frac{\phi(n)}{n!}(-x)^n \end{aligned}

Then:

∫0∞xs−1f(x) dx=Γ(s)ϕ(−s)\int_0^\infty x^{s-1}f(x)\,\mathrm dx=\Gamma(s)\phi(-s)

This is the Ramanujan Master Theorem (Ramanujan CF2100\textcolor{FF8C00}{\text{Ramanujan CF2100}} Theorem). We formally state its proposition:

Ramanujan Master Theorem

For a function f(x)f(x), its Mellin Transform satisfies:

M{f(x)}(s)=∫0∞xs−1f(x) dx=Γ(s)ϕ(−s)\mathcal M\{f(x)\}(s)=\int_0^\infty x^{s-1}f(x)\,\mathrm dx=\Gamma(s)\phi(-s)

if and only if the function f(x)f(x) satisfies:

f(x)=∑n=0∞ϕ(n)n!(−x)nf(x)=\sum_{n=0}^\infty\frac{\phi(n)}{n!}(-x)^n

This theorem describes the general property of the Mellin transform for entire functions.

Applications#

DHL Integral#

First, I will prove the DHL integral mentioned in my blog introduction.
Consider the integral:

I=∫0∞sin⁡xn dxI=\int_0^\infty\sin x^n\,\mathrm dx

Let t=xn,s=1nt=x^n,s=\frac1n, then:

I=s∫0∞ts−1sin⁡t dt=sI∫0∞ts−1eit dt\begin{aligned} I&=s\int_0^\infty t^{s-1}\sin t\,\mathrm dt\\ &=s\mathfrak I\int_0^\infty t^{s-1}e^{it}\,\mathrm dt \end{aligned}

Consider Taylor expanding eite^{it} (I proved in a previous article that this is feasible):

eit=∑n=0∞(it)nn!=∑n=0∞(−t)nn!e−inπ2⏟ϕ(n)\begin{aligned} e^{it}&=\sum_{n=0}^\infty\frac{(it)^n}{n!}\\ &=\sum_{n=0}^\infty\frac{(-t)^n}{n!}\underbrace{e^{-i\frac{n\pi}2}}_{\phi(n)} \end{aligned}

Applying the Ramanujan Master Theorem, we get:

I=I[ sΓ(s)eisπ2]=1nΓ(1n)sin⁡(π2n)I=\mathfrak I[\,s\Gamma(s)e^{i\frac{s\pi}2}]=\frac1n\Gamma\left(\frac1n\right)\sin\left(\frac\pi{2n}\right)

Similarly:

∫0∞cos⁡xn dx=1nΓ(1n)cos⁡(π2n)\int_0^\infty\cos x^n\,\mathrm dx=\frac1n\Gamma\left(\frac1n\right)\cos\left(\frac\pi{2n}\right)

An Anomalous Integral#

Consider the integral:

I=∫0∞dx1+xnI=\int_0^\infty\frac{\mathrm dx}{1+x^n}

Let t=xn,s=1nt=x^n,s=\frac1n, then we have:

∫0∞dx1+xn=s∫0∞ts−11+t dt\int_0^\infty\frac{\mathrm dx}{1+x^n}=s\int_0^\infty\frac{t^{s-1}}{1+t}\,\mathrm dt

Note that:

11+t=∑n=0∞(−t)n=∑n=0∞(−t)nn!Γ(1+n)⏟ϕ(n)\frac1{1+t}=\sum_{n=0}^\infty(-t)^n=\sum_{n=0}^\infty\frac{(-t)^n}{n!}\underbrace{\Gamma(1+n)}_{\phi(n)}

Applying the Ramanujan Master Theorem, we get:

I=sΓ(s)Γ(1−s)=πnsin⁡πnI=s\Gamma(s)\Gamma(1-s)=\frac{\pi}{n\sin\frac\pi n}

First, recall the series expansion form of the ψ\psi function:

ψ(n)=−γ+∑k=0∞(1k+1−1k+n)\psi(n)=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+n}\right)

Rearranging gives:

Hn=ψ(1+n)+γH_n=\psi(1+n)+\gamma

Consider the integral:

I=∫0∞xs−1ln⁡(1+x)1+x dx,R(s)∈(0,1)I=\int_0^\infty\frac{x^{s-1}\ln(1+x)}{1+x}\,\mathrm dx,\mathfrak{R}(s)\in(0,1)

Note that:

ln⁡(1+x)=−∑n=0∞(−x)n+1n+1,11+x=∑n=0∞(−x)n\ln(1+x)=-\sum_{n=0}^\infty\frac{(-x)^{n+1}}{n+1},\frac1{1+x}=\sum_{n=0}^\infty(-x)^n

Then:

ln⁡(1+x)1+x=−∑n=0∞xn+1n+1⋅∑n=0∞(−x)n=−∑n=0∞(−x)n∑k=1n1k=−∑n=0∞Hn(−x)n=−∑n=0∞[ψ(1+n)+γ](−x)n\begin{aligned} \frac{\ln(1+x)}{1+x}&=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\cdot\sum_{n=0}^\infty(-x)^n\\ &=-\sum_{n=0}^\infty(-x)^n\sum_{k=1}^n\frac{1}k\\ &=-\sum_{n=0}^\infty H_n(-x)^n\\ &=-\sum_{n=0}^\infty \left[\psi(1+n)+\gamma\right](-x)^n \end{aligned}

Using the same technique as the previous problem, we obtain:

I=−πsin⁡πs[ψ(1−s)+γ]I=-\frac\pi{\sin \pi s}\left[\psi(1-s)+\gamma\right]

Zeta Function#

Finally, we give the connection between the Riemann Zeta function and Bernoulli numbers.
First:

ζ(s)Γ(s)=∫0∞xs−1ex−1 dx\zeta(s)\Gamma(s)=\int_0^\infty\frac{x^{s-1}}{e^x-1}\,\mathrm dx

Note that the right side is a Mellin Transform. According to the Ramanujan Master Theorem:

1ex−1=∑n=0∞ζ(−n)n!(−x)n\frac1{e^x-1}=\sum_{n=0}^\infty\frac{\zeta(-n)}{n!}(-x)^n

Note that this is very similar to the generating function of Bernoulli numbers, so:

∑n=0∞Bnxnn!=∑n=0∞(−1)nζ(−n)n!xn+1\sum_{n=0}^\infty\textcolor{red}{B_n}\frac{x^n}{n!}=\sum_{n=0}^\infty (-1)^n\frac{\zeta(-n)}{n!}x^{n+1}

Consider unifying the powers. Note that lim⁡n→∞nn!=0\displaystyle\lim_{n\to\infty}\frac{n}{n!}=0, so:

∑n=0∞(−1)nζ(−n)n!xn+1=∑n=0∞(−1)n+1nζ(1−n)xnn!\sum_{n=0}^\infty (-1)^n\frac{\zeta(-n)}{n!}x^{n+1}=\sum_{n=0}^\infty\textcolor{red}{(-1)^{n+1}n\zeta(1-n)}\frac{x^n}{n!}

Comparing coefficients, we get:

ζ(1−n)=(−1)n+1Bnn\zeta(1-n)=(-1)^{n+1}\frac{B_n}{n}

Noting that the odd-indexed Bernoulli numbers (except the first term) are all 00, the formula can be rewritten as:

ζ(1−2n)=−B2n2n,n∈Z\boxed{\zeta(1-2n)=-\frac{B_{2n}}{2n},n\in\mathbb Z}
Ramanujan Master Theorem
https://peakyi.github.io/posts/rmt/ramanujan-master-theorem/
Author
Hiiragi Kagami
Published at
2025-10-23
License
CC BY-NC-SA 4.0
Some Useful Transform
© 2025 Hiiragi Kagami. All Rights Reserved. / RSS / Sitemap
Powered by Astro & Fuwari