hello,
can someone help me to find a solution of the following integration:
integration (0,inf)((sinx)^2/(x^4))
tnx in advence
I don't think your integral converges. The problem is undoubtedly near the origin, since the $\displaystyle x^{4}$ in the denominator behaves very nicely as you go out to infinity, and the trig function is bounded. Near the origin, however, the integrand looks like $\displaystyle 1/x^{2},$ and the integral
$\displaystyle \displaystyle\int_{0}^{1}\frac{1}{x^{2}}\,dx$ diverges.
Are you sure it isn't an $\displaystyle x^{2}$ in the denominator?
I'm not sure evaluating this integral is going to be very productive for you. It doesn't converge, period. Unless you change the limits so that the lower limit is strictly greater than zero, there's not much point in continuing. After all, if we take a finite interval and examine the integral there, we'd have:
$\displaystyle \displaystyle\int_{0}^{\pi/2}\frac{\sin^{2}(x)}{x^{4}}\,dx\ge\int_{0}^{\pi/2}\frac{(\frac{2x}{\pi})^{2}}{x^{4}}\,dx=\frac{4}{ \pi^{2}}\int_{0}^{\pi/2}\frac{1}{x^{2}}\,dx=\infty.$
Does this integral show up in the context of a larger problem? If so, could you please state the original problem?
thanks a lot for your kind reply.
Let me tell you how i came to this equ. i have started with a phase error variance of the RF of a laser which equals to Phi= int(0, inf) (del_v/(pi*f^2))df
where, del_v=laser linewidth.
Now this RF signal is passed through the matched filter with a transfer function of H(f)=(sin(pi*f*T)/(pi*f*T)).
now the error variance becomes= Phi*[H(f)]^2
if i put the corresponding values of Phi and H(f), assuming that pi*f*T=x, i end up on
del_v*T*int(0,inf) [sin(x)^2/x^4]. here i can apply the filter limit as -B/2 to B/2.
Unfortunately, the result still comes as infinity.
What do you suggest?
Hmm. RF is an immensely complicated subject, and I don't have any background at all in it. All I can say is that "phase error variance" sounds like it has to be finite. However, the integral you're ending up with is infinite. Therefore, I would deduce that you're making an error in there somewhere in your derivation of the integral.