Evaluation of double integral

Evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left(\frac{1-e^{-xy}}{xy}\right)^2e^{-x^2-y^2}\text{d}x\text{d}y$$

Solution. One trivial observation is that $$\left(\frac{1-e^{-xy}}{xy}\right)^2=\sum_{n=0}^{\infty}\frac{2(-1)^n(2^{n+1}-1)}{(n+2)!}(xy)^n$$
Moreover  $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(xy)^{2n}e^{-x^2-y^2}\text{d}x\text{d}y=\left(\int_{-\infty}^{\infty}x^{2n}e^{-x^2}\text{d}x\right)^2=\frac{(2n)!^2}{2^{4n}(n!)^2}\pi$$ and $0$ when the power of $xy$ is odd.
Combining the previous two we arrive at the following series $$S=\sum_{n=0}^{\infty}{2n\choose n}\frac{2^{2-2n}-2^{1-4n}}{(2n+1)(2n+2)}\pi$$
Now it is natural to consider the series $$f(x):= \sum_{n=0}^{\infty}{2n\choose n}\frac{x^{2n}}{(2n+1)(2n+2)}$$
Observe that $$(x^2f(x))''= \sum_{n=0}^{\infty}{2n\choose n}x^{2n}\ \ \  (*)$$
Recall that similar central binomials occur at the expansion of  $(1-x)^{-1/2}$ and it indeed turns out that $$\sum_{n=0}^{\infty}{2n\choose n}x^{2n}=\frac{1}{\sqrt{1-4x^2}}$$
Now integrating $(*)$ twice and adjusting the constants we arrive at $$f(x)=\frac{-1+\sqrt{1-4x^2}+x\arcsin(2x)}{4x^2}$$
It remains to calculate $S$ through $f$ - $$S=\pi\left(4f\left(\frac{1}{2}\right)-2f\left(\frac{1}{4}\right)\right)=\frac{4}{3}\pi(3-3\sqrt{3}+\pi)$$

At the last point Abel's theorem for evaluating a function at the end of the interval of convergence of its series is needed. Also some other convergence issues should be made rigorous.

Comments

Post a Comment

Popular posts from this blog

Basel problem type sum, proposed by prof. Skordev

Evaluate $$\sum_{n=1}^{\infty}\frac{1}{2^n n^2}.$$ Proof. The function $$S(x):=\sum_{n=1}^{\infty}\frac{x^n}{ n^2}$$ is well defined for $x\in[-1,1]$ and thus we need to find $S(1/2)$. Using differentiation, we observe that $$S(x)=-\int_0^x\frac{\log(1-t)}{t}\text{d}t.$$ If we make change of variables in the latter integral $t\to 1-t$ we obtain $$S(x)=-\int_{1-x}^1\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (1)$$ On the other hand, using integration by parts, we obtain $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log t\log(1-t)\Big|_{t=0}^x-\int_0^x\frac{\log t}{1-t}\text{d}t.$$ Notice that $\displaystyle \lim_{t\to 0}\log t \log(1-t)=0$ ($\log (1-t)\sim t$ and then L'Hopital), hence $$S(x)=-\int_0^x\log(1-t)\text{d}\log t=-\log x\log(1-x)-\int_0^x\frac{\log t}{1-t}\text{d}t. \ \ \ \ \ (2)$$ Plugging $x=1/2$ and equating $(1)$ to $(2)$ we obtain $$\int_{1/2}^1\frac{\log(1-t)}{t}\text{d} t-\int_{0}^{1/2}\frac{\log(1-t)}{t}\text{d} t= \log\left(\frac{1}{2}\right)^2.$$ On the ot
Read more

Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_{n\to\infty}x_n=0.$ $\textbf{
Read more

A problem proposed by prof. Babev

Consider the sequence $$a_n=\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t.$$ It could be proven that $$\lim_{n\to \infty}n a_n=\frac{1}{2}$$  Now we are going to prove that $$\lim_{n\to\infty}n\left(na_n-\frac{1}{2}\right)=-\frac{1}{4} $$ and show how to derive all such limits. $\textbf{Proof.}$ Since $\displaystyle\int_{n}^{n+1}\sin^2(\pi t)\text{d}{t}=\frac{1}{2}$ (for any $n\in \mathbb{N}$) we can rewrite the limit in question as $$\lim_{n\to \infty}n\left(n\left(\int_{n}^{n+1}\frac{\sin^2(\pi t)}{t}\text{d}t-\int_{n}^{n+1}\frac{\sin^2(\pi t)}{n}\text{d}t\right)\right)=\lim_{n\to \infty}n^2\left(\int_{n}^{n+1}\sin^2(\pi t)\left(\frac{1}{t}-\frac{1}{n}\right)\text{d}t\right)$$ $$=\lim_{n\to \infty}n\left(\int_{n}^{n+1}\sin^2(\pi t)\frac{n-t}{t}\text{d}t\right)$$ Changing the variables $t\to n+s$ and using periodicity of sine, the last simplifies to $$ \lim_{n\to \infty}n\int_{0}^{1}\sin^2(\pi s)\frac{-s}{n+s}\text{d}s$$ Since for $s\in[0,1]$, $\displaystyle \frac
Read more