I'm trying to write a subroutine (in Fortran) to compute integrals of the form $$I=\int_{-L}^{L} f(x)g(y-x) \:\mathrm{d}x, $$
using the convolution theorem and fast Fourier transforms. In my routine, I have discretized $f$ and $g$ uniformly on a domain $(-L,L]$. I then try to compute the integral via $$ I = \mathrm{ifft}(\:\mathrm{fft}(f(x))\cdot\mathrm{fft}(g(x))). $$ However, I cannot get my routine to give me the correct results. As a test, I try to compute $$\int_{-\pi}^{\pi}\mathrm{sin}(x)\mathrm{cos}(y-x)\:\mathrm{d}x, $$ which I believe should give me $\pi\; \mathrm{sin}(y)$. I discretize $x$ into $n$ points. Following the advice of some textbooks, I extend the arrays which contain my discretized functions, and pad them with zeros, so that the sin$(x)$ "signal" looks like: 
and the cos$(x)$ "signal" looks like: 
where the function has also been "wrapped around" as well as zero padded. These two signals are then multiplied together, then multiplied by the grid spacing, $2\pi/n$, and then inverse transformed. The result, (which should be the convolution) looks like this: 
My textbook tells me that part of the result is garbage, which should be discarded, leaving me with the convolution. Looking at the plot above, it looks like I should discard the last half of the signal. The first half of the signal resembles the $\pi$sin$(x)$ that I'm looking for, but it is not scaled correctly. How can I get the correct convolution? I've been stuck on this for some days now and would really appreciate any comments or tips. Thanks.
Edit : Update
I am now trying without the zeropadding. However I am still using "wrap-around" format for the second function. That is, the first half of the array contains the function at positive $x$, and the second half of the array contains the function at negative $x$. The result I get is a sin wave, but still not scaled correctly. In fact here is a plot of the error of my routine for different $n$: 
Clearly its crap.
Edit 2
By extending the arrays to include multiple periods of the signals - as suggested by @copper.hat - instead of zero padding, I am getting good results. My routine now accurately approximates the integral, giving me accuracy to within machine precision as desired : 