Solve Laplace Transform of e^-te^tcost

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Homework Statement



$$L\{ { e }^{ -t }*{ e }^{ t }cost\}$$

Homework Equations





The Attempt at a Solution



$$L\{ { e }^{ -t }*{ e }^{ t }cost\} \\ =L\{ \int _{ 0 }^{ t }{ { e }^{ -\tau }{ e }^{ t-\tau }cos(t-\tau )d\tau } \} \\ =\frac { L\{ { e }^{ t }cost\} }{ s } \\ =\frac { s-1 }{ s[{ (s-1) }^{ 2 }+1] }$$
 
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Unfortunately it isn't correct. For convolution we have:

##L\{f*g\}=L\{f\}\cdot L\{g\}##
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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