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4

Here comes a big hint: Start with the expansions $$ \begin{aligned} e^{xt/2}&=1+\frac{xt}{2}+\frac{1}{2!}\Bigl(\frac{xt}{2}\Bigr)^2+\frac{1}{3!}\Bigl(\frac{xt}{2}\Bigr)^3+\cdots\quad\text{and}\\ e^{-x/2t}&=1-\frac{x}{2t}+\frac{1}{2!}\Bigl(\frac{x}{2t}\Bigr)^2 -\frac{1}{3!}\Bigl(\frac{x}{2t}\Bigr)^3+\cdots. \end{aligned} $$ Multiply them, and look ...


3

Extending what Zach466920 said: $$\partial_t[k(t,n)]=F(n+1)\cdot k(t,n+1)\tag1$$ We can develop a power series solution for this. Let $$k(t,n)=\sum_{m=0}^\infty \kappa(n, m)\ t^m$$ We equation $(1)$ as: $$\sum_{m=1}^\infty m\ \kappa(n, m)\ t^{m-1} =\sum_{m=0}^\infty(m+1)\ \kappa (n, m) \ t^m =F(n+1)\cdot\sum_{m=0}^\infty \kappa(n+1, m)\ t^m$$ Which gives ...


2

We have: $$G(s)=F(s-2) = \frac{1}{3s-2+\sqrt{s^2-4}} $$ so, in order to tackle this problem like the previous one, we should find the coefficients of the Taylor series at $x=0$ of: $$ H(x) = G\left(\frac{1}{x}\right) = \frac{x}{3-2x+\sqrt{1-4x^2}}=\sum_{n\geq 1}g_n x^n\tag{1} $$ to deduce (have also a look at Ramanujan's master theorem): $$ (\mathcal{L}^{-1}...


1

Hint. We have $$ \frac{\partial}{\partial x} Γ(s,x)=-x^{s-1}e^{-x} $$ then, by the chain rule, we get $$ \begin{align} \frac{d}{d x}\left(Γ(1+d,A-c \ln x)\right)&=\frac{-c}{x}\cdot\left.\frac{\partial}{\partial t} Γ(s,t)\right|_{(s,t) =(1+d,A-c \ln x)} \\\\&=\frac{c}{x}\cdot(A-c \ln x)^de^{-(A-c \ln x)} \\\\&=c\:x^{c-1}e^{-A}(A-c \ln x)^d. \end{...


1

$S(x)$ is an entire function, it does not have any logarithmic singularity: $$ S(x)=\sum_{n\geq 1}\frac{x^n}{n\cdot n!}=\int_{0}^{x}\frac{e^t-1}{t}\,dt \tag{1}$$ also since $\frac{e^t-1}{t}$ is an entire function. The RHS of $(1)$ clearly depends on the exponential integral and clearly is not an elementary function.


1

By replacing $t$ with $e^{i\theta}$, we want to prove the identity: $$ e^{ix\sin\theta} = \sum_{n\in\mathbb{Z}} J_n(x)\,e^{in\theta} \tag{1}$$ that (by Fourier inversion) is equivalent to proving that: $$ J_n(x) = \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ix\sin\theta-n\theta\right)\,d\theta. \tag{2}$$ The fastest way is probably to notice that both your series ...


1

Finally, here is the solution. Step 1: Expressing the product of two Meijer G function using an identity from http://functions.wolfram.com/07.34.16.0003.01 \begin{align} G_{2,2}^{1,2}\left(x\Bigg\vert\begin{matrix}1,1\cr 1,0\end{matrix}\right)G_{1,2}^{2,0}\left(2\alpha\sqrt{ab}x\Bigg\vert\begin{matrix}1/2\cr v,-v\end{matrix}\right) = G_{0,0:2,2:1,2}^{0,0:...



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