# Laplace transform of integral

Find the Laplace Transform of:

$$\int_{0}^{t} \frac{Y(u)}{\sqrt{t-u}}du$$

I understand that $$\mathcal{L}\{\int_{0}^{t} Y(u) \ du\} = \frac{y(s)}{s}$$, but I don't understand how this works when other variables are involved (in this case how do we handle the $$\frac{1}{\sqrt{t-u}}$$ term?

## 1 Answer

You are working with semi-primitives, nice! Well, the integral $$\int_{0}^{t}\frac{Y(u)}{\sqrt{t-u}}\,du$$ is the convolution between $$Y(u)$$ and $$\frac{1}{\sqrt{u}}$$, so its Laplace transform is simply the product between $$(\mathcal{L}Y)(s)$$ and $$\mathcal{L}\left(\frac{1}{\sqrt{u}}\right)(s)=\frac{\sqrt{\pi}}{\sqrt{s}}.$$ This is related to fractional calculus since a possible (but not very common) definition of a semi-primitive is $$(D^{-1/2} f)(x) = \mathcal{L}^{-1}\left[\frac{1}{\sqrt{s}}\cdot(\mathcal{L} f)(s)\right](x).$$ In these terms $$\int_{0}^{t}\frac{Y(u)}{\sqrt{t-u}}\,du = \sqrt{\pi}\,(D^{-1/2} Y)(t).$$ It is interesting to check what happens by taking $$Y$$ as a Legendre or Chebyshev polynomial/function, but I do not want to spoil too much of my future work.

• Of course, how could I have forgotten about convolutions! Thanks, mate. – Dr.Doofus Jan 27 at 2:48
• @Dr.Doofus: you're very welcome. – Jack D'Aurizio Jan 27 at 2:49