Laplace transform of convolution with derivative in it

tjosan
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Homework Statement



Hi,

I am wondering how to Laplace transform this expression

f(t)=\int^{\tau}_{0} g(\tau)f'(t-\tau)d\tau
or more precisely
f(t)=\int^{\tau}_{0} sin(8\tau)f'(t-\tau)d\tau

The f'(t-\tau) gets me confused.

Homework Equations



\int^{\tau}_{0} f(t-\tau)g(\tau)d\tau
and the laplace transform of that is:
F(s)G(s)

The Attempt at a Solution


I have no idea how to proceed.

Maybe
F(s)=8/(s^2+8^2)(sF(s)-f(0))
 
Last edited:
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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