How to integrate this?
Evaluate lim
I've had difficulty in using L'Hopital rule. At the same time failed to really understand how to differentiate or evaluate the limit of the x\sin^3x and \int_0^x(x-t)\sin t^2\ dt
Would like to appreciate your help
Answer
Use repeatedly \sin u=u\,{\rm sinc}(u), whereby \lim_{u\to0}{\rm sinc}(u)={\rm sinc}(0)=1. We have \int_0^x (x-t)\sin(t^2)\>dt=x^4\int_0^1(1-\tau)\,\tau^2{\rm sinc}(x^2\tau^2)\>d\tau and \>x\sin^3 x=x^4\>{\rm sinc}^3(x)\ , so that {\int_0^x (x-t)\sin(t^2)\>dt \over x\sin^3 x}={\int_0^1(1-\tau)\,\tau^2\bigl(1+r(x,\tau)\bigr)\>d\tau\over 1+\bar r(x)}\ , whereby \lim_{x\to0}r(x,\tau)=0 uniformly in \tau, and \lim_{x\to0}\bar r(x)=0 as well. It follows that the limit in question is \int_0^1(1-\tau)\,\tau^2\>d\tau={1\over12}\ .
No comments:
Post a Comment