Monthly Archives: January 2018

Ramanujan’s Master Theorem

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“Ramanujan’s Master Theorem”. What a name right? Well it’s also a pretty cool and useful theorem. It says that if f(x) can be expanded as a power series such as this

\displaystyle f(x)=\sum_{n=0}^{\infty} \frac{\phi(n)}{n!}(-x)^n

Then its Mellin Transform can be calculated as follows

\displaystyle \mathcal{M}[f](s)=\int_0^{\infty}x^{s-1}f(x)\,dx=\Gamma(s)\phi(-s)

We can use this formula to calculate the scary-looking definite integral:

\displaystyle I=\int_0^{\infty} e^{-x^3}\sin(x^3)\,dx

First, make the change of variable u=x^3, dx=\frac{1}{3}x^{-2/3}. We then have

\displaystyle I=\frac{1}{3}\int_0^{\infty} e^{-u}\sin(u)u^{-2/3}\,du

Or

\displaystyle I=\frac{1}{3}\Im\int_0^{\infty} e^{-u}e^{iu}u^{-2/3}\,du=\frac{1}{3}\Im\int_0^{\infty} e^{-u(1-i)}u^{-2/3}\,du

Doesn’t this remind you of something? If we let f(u)=e^{-u(1-i)} and s=1/3, we can apply Ramanujan’s Master Theorem.

Remembering the Maclaurin expansion of the exponential,

\displaystyle e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}

Letting z=-u(1-i),

\displaystyle e^{-u(1-i)}=\sum_{n=0}^{\infty}\frac{(1-i)^n}{n!}(-u)^n

And no need to worry about convergence issue since the exponential converges absolutely.  However, we would like simplify that complex number into another form that is easier to work with in this context. Using Euler’s formula, we know that

\displaystyle \sqrt{2}e^{-i\pi/4}=1-i

Thus

\displaystyle e^{-u(1-i)}=\sum_{n=0}^{\infty}\frac{\left(\sqrt{2}\right)^n e^{-in\pi/4}}{n!}(-u)^n

This means that our integral is

\displaystyle I=\frac{1}{3}\Im \left(\Gamma\left(\frac{1}{3}\right)\,\phi\left(\frac{1}{3}\right)\right)

Where

\displaystyle \phi(n)=\left(\sqrt{2}\right)^n e^{-in\pi/4}

Continuing,

\displaystyle \phi\left(-\frac{1}{3}\right)=2^{-1/6}e^{i\pi/12}

Combining everything, we get that

\displaystyle I=\Gamma\left(\frac{4}{3}\right)\!\!\sqrt[-6]{2}\,\sin\frac{\pi}{12}

Thus,

\displaystyle \int_0^{\infty} e^{-x^3}\sin(x^3)\,dx=\Gamma\left(\frac{4}{3}\right)\!\!\sqrt[-6]{2}\,\sin\frac{\pi}{12}

 

Intégrale #4

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Aujourd’hui, je vous propose de découvrir cette intégrale:

\begin{aligned} I&=\int\limits_0^{\infty} \frac{\log(x)}{1+e^x}\,dx \end{aligned}

A première vue, cette intégrale a l’air assez dur. Comme d’habitude, il y a une technique qui vous permet de l’évaluer relativement facilement. Considérons d’abord l’intégrale suivante:

\displaystyle I(s)=\int\limits_0^{\infty} \frac{x^s}{1+e^x}\,dx

Remarquez que nous cherchons à évaluer I'(0). Commençons:

\begin{aligned} I(s)&=\int\limits_0^{\infty} \frac{x^s}{1+e^x}\,dx\\ I(s)&=\int\limits_0^{\infty} x^s e^{-x}\frac{1}{1+e^{-x}}\,dx \end{aligned}

Maintenant, on va développer la fraction en série géométrique:

\begin{aligned} I(s)&=\int\limits_0^{\infty}x^se^{-x}\sum_{n=0}^{\infty} (-1)^n e^{-nx}\,dx\\ I(s)&=\sum_{n=0}^{\infty} (-1)^n \int\limits_0^{\infty}x^s e^{-x(n+1)}\,dx\\I(s)&=\sum_{n=1}^{\infty} (-1)^{n-1} \int\limits_0^{\infty}x^s e^{-nx}\,dx \end{aligned}

Substituons u=nx, et donc du=ndx

\begin{aligned} I(s)&=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s+1}}\int\limits_0^{\infty} u^s e^{-u}\,du \\ I(s)&=\eta(s+1)\Gamma(s+1) \end{aligned}

Où \eta(s) est la fonction eta de Dirichlet, et \Gamma(s) est la fonction Gamma.

Prenant des dérivées,

\begin{aligned} I'(s)&=\eta'(s+1)\Gamma(s+1)+\eta(s+1)\psi(s+1)\Gamma(s+1) \\ I'(0)&=\eta'(1)\Gamma(1)+\eta(1)\psi(1)\Gamma(1)\\I'(0)&=\log(2)\gamma+\frac{1}{2}\log^2(2)-\log(2)\gamma\\I'(0)&=\frac{1}{2}\log^2(2) \end{aligned}

Ce qui veut donc dire que

\displaystyle \int\limits_0^{\infty} \frac{\log(x)}{1+e^x}\,dx=\log^2\left(\sqrt{2}^{\sqrt{2}}\right)

Integral #3

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In today’s post, I have here a real gem for integral aficionados. It truly is marvelous. Okay, let’s get started:

\displaystyle I=\int_{\mathbb{R}} \log\left(1+e^{-x^2}\right)\,dx

Here, \log denotes the natural logarithm. Not only is it much more elegant than using \ln, but it is also standard convention in higher mathematics. Alright, so how de we solve this daunting integral? Well, there are several ways. I’ll show you two of them. And actually, both of them involve series expansion, but one is a bit more general. The first technique consists of introducing a parameter and differentiating under the integral sign with respect to that parameter. It’s also known as Feynman’s trick. If you’re not familiar with that, I have a post on my blog that talks more about it.

Now, you might ask where should we introduce that parameter? Since we are dealing with a logarithm, let’s try inserting it inside its argument, so that when we differentiate, we get rid of the log. How about this:

\displaystyle I(a)=\int_{\mathbb{R}} \log\left(1+ae^{-x^2}\right)\,dx

And after differentiating with respect to a,

\begin{aligned} I'(a)&=\frac{d}{da}\int_{\mathbb{R}} \log\left(1+ae^{-x^2}\right)\,dx \\ &=\int_{\mathbb{R}} \frac{\partial}{\partial a} \log\left(1+ae^{-x^2}\right)\,dx \\&=\int_{\mathbb{R}} \frac{e^{-x^2}}{1+ae^{-x^2}}\,dx \end{aligned}

We can now expand the denominator as a geometric series:

\begin{aligned} I'(a)&=\int_{\mathbb{R}} e^{-x^2}\sum_{n=0}^{\infty} (-1)^n a^n e^{-nx^2}\mathrm{d}x \\ &=\sum_{n=0}^{\infty}(-1)^n a^n \int_{\mathbb{R}}e^{-x^2(n+1)}\,dx \\ &=\sum_{n=1}^{\infty} (-a)^{n-1} a^{n-1} \int_{\mathbb{R}}e^{-nx^2}\,dx \\ &=\sum_{n=1}^{\infty} (-1)^{n-1} a^{n-1} \sqrt{\frac{\pi}{n}} \\ &=\sqrt{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}a^{n-1}}{\sqrt{n}} \end{aligned}

All we have to do now is integrate the series with respect to a. From the initial condition I(0)=0, there are no constants of integration.

\begin{aligned} I(a)&=\int\sqrt{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}a^{n-1}}{\sqrt{n}}\,da\\ &=\sqrt{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}a^n}{n^{3/2}} \end{aligned}

From our original we want to compute I(1):

\begin{aligned} I(1)&=\sqrt{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3/2}}\\ &=\sqrt{\pi}\,\eta(3/2)  \end{aligned}

Where \eta(s) is the Dirichlet eta function. Using this relation between it and the famous zeta function,

\displaystyle \eta(s)=\left(1-2^{1-s}\right)\zeta(s)

We can finally compute our integral:

\begin{aligned} I(1)&=\sqrt{\pi}\,\eta(3/2)\\&=\sqrt{\pi}\left(1-2^{-1/2}\right)\zeta(3/2)\\&=\sqrt{\pi}\left(1-\frac{\sqrt{2}}{2}\right)\zeta(3/2) \end{aligned}

And that is the best we can do because \zeta(3/2) has no closed form. We thus conclude that

\displaystyle \int_{\mathbb{R}} \log\left(1+e^{-x^2}\right)\,dx=\sqrt{\pi}\left(1-\frac{\sqrt{2}}{2}\right)\zeta(3/2)=1.356187790...

In a follow-up post, I’ll show to you the second technique for computing this integral.