It's nice to see a contribution of mine from almost four years ago come up, as just one of many people who have made contributions to this site and get referenced as well. Besides providing answers to many queries, the site appears to have become a searchable source of information.
The reason that the first conv example works and the second doesn't is due to the different x arrays used to create the pdfs. Suppose the fft transforms y = y(x) to z = z(ilam) (ilam being inverse wavelength, which seems to have no common symbol for it). The fft puts z( ilam=0 ), the DC value, as the first element in the resulting z array. Similarly the fft assumes that y( x=0 ) is the first element in the y array. If x=0 is somewhere else, then z picks up a phase shift in the ilam domain that can lead to incorrect results.
Your first example uses the correct x array and the method works. The second example starts out at a negative value of x to accommodate the gaussian, leading to problems as you can see.
The most straightforward approach to allow negative values of x is to start with an x array with x=0 at the midpoint (element n/2+1 for even n). Define y(x) and then use ifftshift to put the DC y value at y(1). I tend to think of the first array as a plotting array and the shifted array as the real array. After some ffts and then an ifft to get back in the x domain, fftshift up to put y( x=0 ) back in the center. For example,
y = (1/(sqrt(pi)*sig))*exp(-(x-mu).^2/sig^2);
yconv = (1/(sqrt(pi*mp1)*sig))*exp(-(x-mp1*mu).^2/(mp1*sig^2));
z = fft(ifftshift(y)).^mp1*delx^mp1;
yconvfft = fftshift(ifft(z))*N*delilam;
For normalization, since you are using the fft (which only does sums) to approximate an integral, there is a factor of the array spacing del_x for each transform. Coming back there is one ifft so there is a single factor of del_ilam. Except the ifft automatically divides by the number of points N, so you need to multiply by N to make up for that.
I do have one question. In the second example you have 2^13 = 8192 points instead of, say, 10000. Could you let me know the rationale behind using 2^n points?