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Given a convolution integral $$ g(y) =\int_a^b\varphi(y-x)f(x)dx=\int_{-\infty}^{+\infty}\varphi(y-x)f(x)\mathbb{I}_{[a,b]}(x)dx $$ where $\varphi(x)= \frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\...

In the chemical analysis by instruments, the signals of several molecules are overlapped which makes it difficult to determine the true area of each peak, such as those shown in red. I simulated this ...