Convolution Product $\sin t *\sin t$ by complex replacement.

Question

I want to compute

$$\sin t * \sin t=\int_0^t \sin u \sin (t-u) \, d u.$$

I have already tried complex replacement:

$$\int_0^t \sin u \sin (t-u) \, \mathrm d u = Im{\int_0^t e^{iu} e^{i(t-u)}\,d u }=Im\int_0^t e^{it}\, d u .$$

The last expression is obviously not $\sin t * \sin t$, which I was looking for. What did I do wrong? I would be glad at any help.

Answer 1

The mistake here is that $Im(a \cdot b) \neq Im(a) \cdot Im(b)$. For example $e^{i\theta} \cdot e^{-i\theta} = 1$ and so $Im(e^{i\theta} \cdot e^{-i\theta}) = 0 \neq -\sin^2(\theta) = Im(e^{i\theta}) \cdot Im(e^{-i\theta})$.

Instead, expand $\sin(t-u)$ in the integrand as $\sin(t)\cos(u)-\cos(t)\sin(u)$. Which yields $$\sin(t)\int_0^t \sin(u)\cos(u) du - \cos(t) \int_0^t \sin^2(u)du.$$

The first integral can be accomplished by rewriting $\sin(u)\cos(u) = 1/2 \sin(2u)$ and the second can be integrated by changing $\sin^2(u) = 1/2(1-\cos(2u))$.

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If you want to do this with a complex replacement then you can replace $\sin(\theta) = \frac{e^{i\theta}-e^{-i\theta}}{2i}$.

This gives $$\int_0^t \left( \frac{e^{iu}-e^{-iu}}{2i} \right) \left( \frac{e^{i(t-u)}-e^{-i(t-u)}}{2i} \right)du.$$

You can then multiply the numerator out and integrate each exponential one at a time.

Answer 2

Just use the identity

$$\sin(a-b) = \sin a \cos b - \sin b \cos a $$

and things will be straightforward.

Answer 3

@science Please have a look at a very similar post where I apply your suggestion but still do not come up with the right answer. Calculate the convolution of the product of two identical sine functions. (5.6-7)