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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?

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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.

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

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