But that's the whole point! \delta(x) is a distribution thus it does not behaves like ordinary functions. Your point of view is that the integral does not exist, because it can take multiple values, but I say that it can be defined if you choose the value of \theta(0) yielding to...
The answer comes by joining the posts of jostpuur and tiny-tim. :smile:
The second integral is
I=\int_{0}^{+\infty} f(x) \delta (x) dx = \int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx
where \theta(x) is the step function. Thus I equals
I=\theta(0)\,f(0)[/itex]
and the question...
The Bessel functions satisy the recurrence relations
J_{n-1}+J_{n+1}=\frac{2\,n}{x}\,J_n\quad \text{and} \quad J_{n-1}-J_{n+1}=2\,J_n'
Adding these, you get
x\,J_{n-1}=n\,J_n+x\,J_n'\overset{n=2}\Rightarrow x\,J_1=2\,J_2+x\,J_2'
while integrating from 0 to \infty
\int_0^\infty...
Break the integral to two ones one from (-\infty,0),(0,+\infty), transform the first to (0,\infty) and combine them to get
I=\int_0^\infty\frac{\ln(x^2+a^2)}{x^2+1}
Use a semi-circle to include the residue i in order to find I=2\,\pi\ln(a+1).
This equation is a Riccati type equation, i.e. y'(t)=a(t)\,y(t)^2+b(t)\,y(t)+c(t). The general solution can be found if one knows a particular solution y_p(t) by the transformation y(t)=y_p(t)+\frac{1}{u(t)} which makes it a linear one.
For the problem at hand if you write...
The equations a\,x+b\,y=c are called Linear Diophantine Equations. If you acn find a particular solution (x,y)=(x_o,y_o) then you can find the general solution by writting x=x_o+\lambda\,t,\,y=y_o+\mu\,t. Plugging these to the original equation you determinate the values of \lambda,\,\mu...