# quark1245

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 Jun16 accepted Chain relation in three variables Jun16 asked Evaluation of a product of sines Apr15 asked Chain relation in three variables Apr11 accepted Improper integral of $\dfrac{x}{e^x-1}$ Apr10 asked Improper integral of $\dfrac{x}{e^x-1}$ Apr10 awarded Editor Apr10 revised Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$deleted 1 characters in body Apr5 accepted Probability of throwing the same multiset twice in a row with six dice Mar28 comment Probability of throwing the same multiset twice in a row with six diceYes, it helps in the sense that now it is immediate to conclude, but is there some slick way to do this? I mean, how can one produce your table during an examination, when time is limited? (this is part of an exercise taken from an exam paper in probability) Mar28 asked Probability of throwing the same multiset twice in a row with six dice Mar9 accepted Random walk problem Mar8 comment Random walk problemAbout your second hint: the condition we need is that the number of left moves is equal for the two drunks, or that the number of left moves of the first + the number of right moves of the second is equal to N. The number of all possible walks is $4^N$. Call $n$ the number of left moves of the first drunk. We can choose these $n$ moves in $(N\,\,\,\, n)$ ways. The number of left moves of the second drunk must be $n$ too and so the probability is $\sum_{n=0}^N=(N \,\,\,\, n)^2/4^N$, that is $4^{-N}(2N\,\,\,\, N)$. Is this right? Thanks Mar8 comment Random walk problemYes, it is just what I thought and following that idea I found the summation above. Can you rewrite it in a nicer form? Mar8 accepted How to calculate the principal part of improper integral? Mar8 accepted Find locus of points in the plane Mar8 accepted Evaluate improper integral $(\cos(2x)-1)/x^2$ Mar8 asked Random walk problem Feb19 comment Evaluate improper integral $(\cos(2x)-1)/x^2$@GEdgar: oh yes, thanks for pointing that out. Feb19 comment Evaluate improper integral $(\cos(2x)-1)/x^2$@GEdgar: the absolute value along the large semicircle is less then $(2\pi R)$ times the maximum of the absolute value of the integrand, that is $2/R^2$. As $R\to\infty$, the all thing goes to zero... Feb19 comment Evaluate improper integral $(\cos(2x)-1)/x^2$Now we are rescued by the half residue theorem that states that the last integral is $\pi i$ times the residue in zero, which is $2i$. Change the sign because the semicircle is oriented negatively. So, the integral is $-2\pi$. Done. Yet, can you state more precisely where my previous argument goes wrong?