Eva.uate : \int_{-2n}^{2n+\frac 12}{\sin(\pi x)\left\{\frac x2 \right\}} \;\; \mathrm{d}x
1im geting exactly same as aditya,
break it from -2n to 2n and 2n to 2n + 1/2.
put {x} = x-[x]
so we hav 3 integrals out of which the one with [x] is zero. 'cos when u break it into many integrals (as we so generally for any integral involving step function) each term on integration will be 0..
now from by parts use ∫xsinx = sinx - xcosx for the remaining 2 integrals.
71But the answer has no 2 for one term ie., 1Î 2 - 2nÎ .
Aditya, How did you do and it? Also one point to note is that both function are periodic with period 2. Hence job is also more easier.
1do mention if it's your doubt or no. it'll save me time writing the steps.
262calculate the integral from 0 to 2 .
now since the function has a period 2,
given integral = 2n * (integral from 0 to 2) + (integral from 0 to 1/2)
71Hmm thanks aditya/risabh.
@Rishbah. Yes I wished to know the method pursued by u.
1I am also getting the same as aditya, simply by parts. 1st is x/2 and second fn is sinpix
9breaking down the integration
-2n∫2nsin(Πx){0.5x}dx+2n∫2n+0.5sin(πx)sin(Πx){0.5x}dx
first is odd and periodic hence=0 period 2
second 1:
0∫0.50.5xsin(Πx)dx