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6

Not sure if it is your book's problem or what, but the Weierstrass product formula is $$ \frac1{\Gamma(z)} = ze^{\gamma z} \prod_{n=\color{red}1}^\infty \left(1+\frac zn\right)e^{-z/n} $$ (Note that $(1+\frac z0)e^{-z/0}$ is undefined, so the product cannot start from 0.) So after taking inverse square root your solution should be $$ \left| \Gamma(\alpha + ...


4

We consider the Bessel functions \begin{align*} J_p(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}\\ J_{-p}(x)&=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1-p)}\left(\frac{x}{2}\right)^{2n-p}\\ \end{align*} Blue question: Due to the symmetry of $J_p(x)$ and $J_{-p}(x)$ the author seems ...


2

strictly monotone increasing. Any such function is continuous almost everywhere. But what do you mean by singularities? It could have jump discontinuities, in fact, it could have infinitely many. As pointed out by Merlinsbeard, it is in fact almost everywhere differentiable (which is even stronger)


2

Why (3) is so different from (1) ? Because if we rewrite the double integrals as iterated ones, for (1) we get $$\int_0^1 \left(1-x_2\right)^{\alpha_3-1}\left({\color{blue}{\int_0^{x_2}x_1^{\alpha_1-1}\left(x_2-x_1\right)^{\alpha_2-1}dx_1}}\right)dx_2,$$ and for (3) we find $$\int_0^1 x_1^{\alpha_1-1}\left({\color{red}{\int_0^{x_1} ...


2

The trick is to use $$t^2+b^2=(t+i b)(t-ib)$$ and use partial fraction decomposition $$\frac t{t^2+b^2}=\frac{1}{2 (t+i b)}+\frac{1}{2 (t-i b)}$$ So $$\int \frac{t\sin at}{t^2+b^2}dt=\frac{1} 2\int\frac{\sin at}{t+ib}dt+\frac{1} 2\int\frac{\sin at}{t-ib}dt$$ So, now, consider $$I=\int \frac{\sin at}{t+c}dt$$ Change variable $t=u-c$ to get $$I=\int\frac{\sin ...


2

Since you ask multiple questions, I suppose what follows isn't a complete answer. I learned this trick the other day, so I'm more than happy to share it here, If you want an integral that can generalize the recurrence, $$(1) \quad \mathfrak F(0)=1,\quad \mathfrak F(n+1)=\mathfrak F(n)\times F(n+1).\tag3$$ You can do so by assuming the formula is of the ...


2

Let me consider the two integrals $$I = \int \mathrm{e}^{-v}\int_0^{a - bv} \mathrm{e}^{-u^2} du\,dv$$ $$J= \int \int_0^{a + bv} \mathrm{e}^{-u^2} du\,dv$$ First $$\int_0^{a - bv} \mathrm{e}^{-u^2} du =\frac{\sqrt{\pi } }{2} \text{erf}(a-b v)$$ which makes $$I=\frac{ \sqrt{\pi }}{2} \int e^{-v} \text{erf}(a-b v)\,dv$$ This one can be integrated by parts ...


2

It turns out to be an identity about Bernoulli numbers, $B_2=B_{14}-1$. They both are rational numbers, and their denominators have to be the same by the Von Staudt-Clausen theorem. Equality of numerators, too, happens more or less by chance.


1

We have the identity: $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1} $$ hence by taking $a=\frac{N}{2}+1$ and $b=\frac{N}{2}+\frac{1}{2}$ it follows that: $$ \psi\left(\frac{N}{2}+1\right)-\psi\left(\frac{N}{2}+\frac{1}{2}\right)=2\sum_{n\geq 0}\frac{1}{(2n+N+2)(2n+N+1)}\tag{2}$$ but $\sum_{m\geq 0}\frac{1}{(2m+1)(2m+2)}$ is ...


1

Hint: Use Integration by parts, where you integrate $f(x)=e^{-x}$ and differentiate $P_n(x)$, then use the Rodrigues' formula.


1

This can also be done via elliptic integrals and it is possible to get a closed form involving $\Gamma(1/4)$ but the calculations are complicated. We need to use the power series for $\text{dn}(u, k)$ and compare it with its Fourier series. Thus we have $$\text{dn}(u, k) = 1 - k^{2}\frac{u^{2}}{2!} + k^{2}(4 + k^{2})\frac{u^{4}}{4!} - \cdots\tag{1}$$ and the ...


1

hint: $g(x,y) =x+y+x^2+y^2 = 12\implies xy + z^2 = xy + (12-x-y)^2= 3xy + 144+x^2+y^2-24(x+y)=3xy+144+12-25(x+y)=3xy-25(x+y)+156= f(x,y)$. Thus we have: $\nabla f = \lambda \nabla g\implies 3y-25 = \lambda(1+2x), 3x-25=\lambda(1+2y)\implies 3(y-x) = 2\lambda(x-y)\implies x = y$ or $\lambda = -\dfrac{3}{2}$. Can you continue from here?



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