Proof of a Ramanujan Integral

Question

While studying Ramanujan's Collected Papers I came across a paper titled "Some Definite Integrals" which appeared in Messenger of Mathematics, ${\tt XLIV}, 1915, \mbox{10-18}$. It contains lot of weird integrals for which Ramanujan has given proofs.

• However in one instance he discusses about the integral \begin{align} &\int_{0}^{\infty}\frac{dx}{\left(1 + x^{2}\right)\left(1 + r^{2}x^{2}\right)\left(1 + r^{4}x^{2}\right)\cdots} \\[5mm] = &\ \frac{\pi}{2\left(1 + r + r^{3} + r^{6} + r^{10} + \cdots\right)}\label{1}\tag{1} \end{align} where $0 < r < 1$.

• Ramanujan derives this formula from \begin{align} &\int_{0}^{\infty}\frac{\left(1 + arx\right)\left(1 + ar^{2}x\right)\cdots}{\left(1 + x\right)\left(1 + rx\right)\left(1 + r^{2}x\right)\cdots}x^{n - 1}\,\mathrm{d}x \\[5mm] = &\ \frac{\pi}{\sin\left(n\pi\right)} \prod_{m = 1}^{\infty}\frac{\left(1 - r^{m - n}\,\,\right)\left(1 - ar^{m}\,\right)}{\left(1 - r^{m}\,\right)\left(1 - ar^{m - n}\,\,\right)}\label{2}\tag{2} \end{align} where $0 < r < 1, n > 0, 0 < a < r^{n - 1}$ and $n$ is not an integer and $a$ is not of the form $a = r^{p}$ where $p$ is a positive integer.

• Unfortunately, Ramanujan does not prove the formula (\ref{2}).

Is there any direct approach to establish (\ref{1}) without using (\ref{2}) or some way to establish (\ref{2}) $?$.

Answer 1

You can use residues, let $f(z)=\frac{1}{(1+z^2)(1+r^2z^2)(1+r^4z^2)...}$ this has an infite set of singularities at $z=\pm i 1/r^{n},n\in\Bbb N_0$. We can see that $\large\int_0^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx=\frac{1}{2}\int_{-\infty}^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx$

So we will consider the upper semi circle contour, spanning over $-\infty$ to $+\infty$, only the positive singularities are in this contour, i.e. $z=i\frac{1}{r}$.

Now $\operatorname{Res}(f(z),z=\frac{i}{r^{n}})=\large\lim_{z\to i\frac{1}{r^{n}}}(z-i\frac{1}{r^{n}})f(z)=\large\lim_{z\to i\frac{1}{r^{n}}}\frac{1}{r^n}(r^nz-i)f(z)$

$=\lim_{z\to i\frac{1}{r^{n}}}\frac{1}{r^n(r^nz+i)}\prod_{j=0,j\ne n}^\infty\frac{1}{(1+x^{2}r^{2j})}$

$=\large\frac{1}{2ir^n}\prod_{j=0,j\ne n}^\infty\frac{1}{(1-r^{2j-2n})}$

Now $\frac{1}{2}\int_{-\infty}^\infty\frac{1}{(1+x^2)(1+r^2x^2)(1+r^4x^2)...}dx=\frac{1}{2}2\pi i\sum \operatorname{Res}(f)=\frac{1}{2}\pi i\sum_{n=0}^\infty\frac{1}{ir^n}\prod_{j=0,j\ne n}^\infty\frac{1}{(1-r^{2j-2n})}$

From here I'm not sure how to reach the closed form, but hopefully this helps to show a different approach, even if you are not too well versed in integrals by residue :)

Answer 2

I find this relationship charming.

Here is what I have so far as an alternative solution without resorting to complex analysis. For $n=0,1,...$ define \begin{align} I_n=\int_0^{\infty} \frac{dx}{(1+x^2)(1+r^2x^2)...(1+r^{2n}x^2)} \end{align} Doing many partial fractions I could establish for $n=1,2,3,4,...$ that \begin{align} I_{n}&=\frac{1-r^{2n-1}}{1-r^{2n}}I_{n-1}\\ I_0&=\frac{\pi}{2} \end{align} Therefore \begin{align} I_{\infty}&=\frac{\prod_{n=1}^{\infty}\Big(1-r^{2n-1}\Big)}{\prod_{n=1}^{\infty}\Big(1-r^{2n}\Big)}\frac{\pi}{2} \end{align} Now using the same arguments as in this post, we can establish $(1)$.

Answer 3

Ramanujan does indeed give a proof, he shows that the product in your second integral can be expanded as a power series in $x$ and makes use of his master theorem (without comment). This power series is related to the q-binomial theorem (where $\left(ar;r\right)_{n}$ is the q-Pochhammer symbol).

$$\prod_{n=1}^{\infty}\frac{\left(1-axr^{n}\right)}{\left(1-xr^{n-1}\right)} = \sum_{i=0}^{\infty}\frac{\left(ar;r\right)_{i}}{\left(r;r\right)_{i}}x^{i}$$

proof:

(1) Denoting the above product $f(x)$.

$$f(x) = \sum_{i=0}^{\infty}C_{i}x^{i}$$

(2) We can see that $f(x)$ satisfies the functional equation $f(x) = \frac{\left(1-axr\right)}{\left(1-x\right)}f(xr)$, and so...

$$\left(1-x\right)f(x) = \left(1-axr\right)f(xr)$$

and...

$$\sum_{i=0}^{\infty} \left(C_{i}x^{i}-C_{i}x^{i}r^{i}\right) = \sum_{i=0}^{\infty} \left(C_{i}x^{i+1}-aC_{i}x^{i+1}r^{i+1}\right)$$

(3) Re-indexing the sum on the right and equating coefficients we find:

$$C_{i} = \frac{C_{i-1}\left(1-ar^{i}\right)}{\left(1-r^{i}\right)}$$

If we multiply out the original product we can see that $C_{0} = 1$, so, iterating the above expression we find $C_{i} = \frac{\left(ar;r\right)_{i}}{\left(r;r\right)_{i}}$. Since this is a Mellin transform of a power series in $x$, we can use Ramanujan's master theorem and the fact that:

$$\left(x;y\right)_{-n} = \frac{\left(x;y\right)_{\infty}}{\left(xy^{-n};y\right)_{\infty}}$$

Answer 4

I must say, before solving the integral I didn't expect there is such an elementary approach.

We don't have to deal with $q$ binomial theorems or messy partial fractions, just let $f_n(x)$ denote $\displaystyle \frac{1}{(1+x^2)(1+x^2 q^2)\cdots (1+x^2q^{2n-2})}$

and realize that $f_n$ satisfies $$\begin{split} f_n(xq)=& \left(\frac{1+x^2}{1+x^2 q^{2n}}\right)f_n(x) \\ =&\frac{1}{q^{2n}}f_n(x)\left(1-\frac{1-q^{2n}}{1+x^2 q^{2n}}\right) \\ = & \frac{1}{q^{2n}}\left[f_n(x)-(1-q^{2n})f_{n+1}(x)\right] \end{split}$$ So for $\Re (q)>0$: $$(1-q^{2n-1})\int_0^{\infty} f_n(x)\, dx =(1-q^{2n})\int_0^{\infty} f_{n+1}(x)\, dx$$

Which is the relation in @Math-fun 's answer, that immediately solves the integral.

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My approach also works to derive a specific case $a=0$ in $(2)$.

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Takeaway: If you have a function $f(x)$ involving the $q$ shifted factorial, it is probably a good idea to study its relation between $f(xq), f(xq^2)$ etc.

Answer 5

Here is a way you can do it with brute force with the aid of change of variables and Fubini's Theorem. Let \begin{align*} I &= \int_{0}^\infty \frac{1}{(1+x^2)(1+r^2x^2) \ \dots (1+r^{2n}x^2)} \ dx. \end{align*} Note for each $i \in \lbrace 1, \ \dots \ ,n \rbrace,$ we can write \begin{align*} \frac{1}{1+r^{2i} x^2} &= \int_{0}^{\infty} e^{-(1+r^{2i}x^2) y_i} \ dy_i. \end{align*} Thus, using Fubini's Theorem,

\begin{align*} I &=\int_{0}^\infty \int_{0}^\infty \ \dots \ \int_{0}^\infty e^{-y_1(1+x^2)} e^{-y_2(1+r^2x^2)} \ \dots \ e^{-y_{n}(1+r^{2n}x^2)} \ dy_n \ \dots \ dy_1\ dx \\ &=\int_{0}^\infty \int_{0}^\infty \ \dots \ \int_{0}^\infty e^{-(y_1 + \ \dots \ + y_n)} \ e^{-(y_1+r^2y_2 \ \dots \ + r^{2n}y_n)x^2} \ dy_n \ \dots \ dy_1 \ dx \\ &=\int_{0}^\infty \int_{0}^\infty \ \dots \ \int_{0}^\infty e^{-(y_1 + \ \dots \ + y_n)} \ e^{-(y_1+r^2y_2 \ \dots \ + r^{2n}y_n)x^2} \ dx \ dy_n \ \dots \ dy_1. \end{align*} To carry out the integral with respect to $x,$ use the one dimensional change of variables $x=\frac{t}{\sqrt{y_1+r^2y_2 \ \dots \ + r^{2n}y_n}}$ and the well-known Gaussian integral $\int_{0}^\infty e^{-t^2} \ dt= \frac{\sqrt{\pi}}{2}.$ We get

\begin{align*} I &= \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \ \dots \ \int_{0}^{\infty} \frac{e^{-(y_1 + \ \dots \ + y_n)}}{\sqrt{y_1+r^2y_2 \ \dots \ + r^{2n}y_n}} \ dy_n \ \dots \ dy_1. \end{align*}

Now, perform the multidimensional change of variables \begin{align*} y_i &= t_i - r^2 t_{i+1}, \quad 1 \leq i <n \\ y_n &= t_n. \end{align*} By induction on $n,$ it is easy to verify $\left|\frac{\partial(y_1 ,\ \dots \ , y_n)}{\partial(t_1 , \ \dots \ , t_n)}\right|=1$ and that transformed region of integration is

\begin{align*} t_1 &>0 \\ 0 &< t_i < \frac{t_{i-1}}{r^2},\quad 2 \leq i \leq n. \end{align*}

With this, \begin{align*} I &= \frac{\sqrt{\pi}}{2} \int_{0}^{\infty} \int_{0}^{t_1/r^2} \ \dots \ \int_{0}^{t_{n-1}/r^2} \frac{e^{-t_1-(1-r^2)(t_2 + \ \dots \ + t_n)}}{\sqrt{t_1}}\ dt_n \ \dots \ dt_2 \ dt_1. \end{align*}

This makes $I$ an "elementary" integral that can be evaluated either by hand or with Mathematica. To carry out the integration with respect to $t_n, \ \dots \ , t_2,$ one must repeatedly apply the exponential identity $$\int_{0}^{t} e^{-ax} \ dx = \frac{1-e^{-at}}{a}$$ for $a>0,$ which is pretty messy. To carry out the final integration with respect to $t_1,$ make the substitution $t_1=u_1^2$ which will transform the resulting integrand into a Gaussian function.
At the moment, I do not know of a clean, slick way to evaluate this multidimensional integral and arrive at the general answer, but I will look into it.