Recent content by Domdamo

Thank you very much.

Laplace transform of eq. [1] [4] F1(p)-exp{-pi*p}*F1(p) = F2(p) Rearranging eq. [4] [5] F1(p) = frac{1}{1-exp{-pi*p}}*F2(p) Inverse LT of eq. [5]

Dear Ray! Thanks Ray for the information. With your explanation I have understood how to come in the ##\epsilon##. I also got dropped off the email list. (Maybe because of the new face of the physics forums.) That's why I just saw your answer.

Dear Ray, I do not to find out where the ##\epsilon=\text{sign}\left[\mathfrak{Im}(x)\right]## come from, when I evaluated the ##k_{5,7}##. My calculation for ##|x|## large: $$y_5=U[a,c,x]\propto x^{-a}$$ $$y_5^{'}=-a\cdot U[a+1,c+1,x]\propto -a\cdot x^{-a-1}$$ $$y_7=e^x\cdot...

Thanks Ray. :) With this help , I think I am capable to solve my problem.

I did not define my problem clearly: We know from the Abel's theorem that the wronskian of two solutions for the confluent hypergeometic equation equal with this (if ##x\neq 0##): W(y_m,y_n)(x)=\kappa_{m,n}e^{-\int\frac{c-x}{x}dx}=\kappa_{m,n}\cdot x^{-c}\cdot e^{x} where ##\kappa_{m,n} ##...

According to [Erdely A,1953; Higher Transcendental Functions, Vol I, Ch. VI.] the confluent hypergeometric equation \frac{d^2}{d x^2} y + \left(c - x \right) \frac{d}{d x} y - a y = 0 has got eight solutions, which are the followings: y_1=M[a,c,x] y_2=x^{1-c}M[a-c+1,2-c,x]...