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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...

А smoker has two matchboxes in his pocket, each containing $n$ mathes. When he wants to smoke, he selects randomly one of the boxes, and takes a match from it. At some point he finds out, that one of the boxes is empty. What is the probability that in the other box there are exactly $r$ matches left (for $r\in\{0,1,\ldots,n\}$). What about the expected number of the matches in the other box.   If $X$ is the random variable "number of matches in the other box", then one directly finds that $$\text{P}(X=r)={2n-r \choose n-r}\frac{1}{2^{2n-r}}$$  Now we will evaluate $$\text{E}X=\sum_{r=0}^{n}r{2n-r \choose n-r}\frac{1}{2^{2n-r}}$$  The sum of all probabilities being $1$ means here that $$\sum_{r=0}^{n}{2n-r \choose n-r}\frac{1}{2^{2n-r}}=1$$ Change the variables $r=n-d$ in the last and and we obtain $$\sum_{d=0}^{n}{n+d \choose d}\frac{1}{2^{n+d}}=1 \ \ \ \ \ \ \  (*)$$ Make the same change in the sum for the expectation $$\text{E}X=\sum_{d=0}^{n}(n-d)...