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Take the Laplace transform of

$$ \int_{0}^{t}x^2(x-t)^4 \cos(x)dx .$$

I'm not quite sure where to start...

Mhenni Benghorbal
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Jay
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2 Answers2

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Hint: Use the convolution property for the Laplace transform

$$\mathcal{L}(f*g) =\mathcal{L}(f)\mathcal{L}(g). $$

See this related technique.

Note: You can take $f(x)=x^4 $ and $g(x)=x^2\cos(x)$. Finding the Laplace transform of $x^4$ is easy. For $g(x)=x^2\cos(x)$ you can use the Laplace property

$$ \mathcal{L}(x^nh(x)) = (-1)^n H^{(n)}(s), $$

where $H(s)$ is the Laplace transform of $h(x)$.

Community
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Mhenni Benghorbal
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  • Okay. So the Laplace of x^4 would be 4!/s^5. Cool. So that would mean that I have to take the second derivative of x^2cos(x)? To do this I would change it to be -1(d/ds)L(h(x)) where h(x) is cos(x)? – Jay Apr 17 '14 at 04:35
  • @Jay: To find Laplace transform of $x^2\cos(x)$, you need to differentiate twice the the Laplace transform of $\cos(x)$. – Mhenni Benghorbal Apr 17 '14 at 05:18
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I had not found any other way than to follow the above suggested. So I just did the labor calculation.

From $$\mathcal{L}(f*g) =\mathcal{L}(f)\mathcal{L}(g),$$ we can set

$f(x)=x^2\cos(x)$ and $g(x)=x^4$ and

Since $\mathcal{L}[(-t)^nf(t)]=F^{(n)}(s)$, we can find that $$\mathcal{L}[t^2\cos(t)]=\frac{d^2}{ds^2}\frac{s}{s^2 + 1}=\frac{2s(s^2-3)}{(s^2 + 1)^3}$$ and together with $\mathcal{L}[t^4]=\frac{4!}{s^5}$, the final answer will be $$ \mathcal{L}[x^2\cos(x)*x^4]=\frac{2s(s^2-3)}{(s^2 + 1)^3}\frac{4!}{s^5}=\frac{48(s^2-3)}{s^4(s^2 + 1)^3}. $$

MathArt
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