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Yes, you can do it directly using integration by parts to get to $(x+y)B(x+1,y)=x B(x,y)$. With this, you then actually want to go the other way: write the formula as $$ B(x,y) = \frac{x-1}{x+y-1} B(x-1,y). $$ Now you iterate/use induction to get to the formula $$ B(m,n) = \frac{(m-1)(m-2) \dotsm (m-(m-1))}{(m+n-1)(m+n-2)\dotsm (m+n-(m-1))} B(0,n) = ...


2

No It is not In terms of elementary functions, But we can write in Infinite Series form. Using $$\displaystyle \sin x= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+..........$$ So $$\displaystyle \frac{\sin x}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{4!}-\frac{x^6}{6!}.........$$ So $$\displaystyle \int \frac{\sin x}{x}dx = \int ...


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For $\epsilon\leq 1/e$, $$T \geq e^{-W_{-1}(-\epsilon)}=-\epsilon^{-1}W_{-1}(-\epsilon)$$ where $W_{-1}$ is the second branch of the Lambert $W$ function. The Lambert $W$ functions are branches of the inverse of $f(z)=ze^z$. It cannot be expressed in terms of elementary functions, so this is the best you can do. Answer courtesy Wolfram Alpha.



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