Recent content by Rainbow Child

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 standard Gaussian integral which cepheid stated is \int_{-\infty}^\infty e^{-y^2}\,d\,y=\sqrt\pi

Oupps! VietDao29 was faster :smile:

Try this x=2\,\arctan u\Rightarrow u=\tan \frac{x}{2},\,d\,x=\frac{2\,du}{1+u^2} with \sin x=\frac{2\,\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\Rightarrow \sin x=\frac{2\,u}{1+u^2}

The first one reduces to the second one by the change of variables x=\frac{1}{2}\,(v+u),\,y=\frac{1}{2}\,(v-u)

leads to (x^\alpha\,y)'=2\,x\,\sin x

Do you know the derivative of g(x)=a\,x^n ?

Write it as \frac{d\,P}{P\,(a-b\,\ln P)}=d\,t and set \ln P=u

You are welcome! :smile: Just edit the title and marked it [SOLVED]

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...

Since x\in(0,\infty)\Rightarrow |x|=x. Now combine the two integrals.