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How to find the Laplace transform of the following function: $$f(t)=t e^t$$ $$F(s)=\int_0^\infty (te^t e^{-st})dt$$ What method do I use to find the integral?

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  • $\begingroup$ Try a computer algebra program, like wxMaxima. $\endgroup$
    – avs
    May 17, 2017 at 1:30
  • $\begingroup$ integration by parts (combine the e-powers first, remember, you are integrating w.r.t. $x$, not $s$) $\endgroup$
    – imranfat
    May 17, 2017 at 1:31
  • $\begingroup$ Combine the exponentials and use integration by parts $\endgroup$
    – Triatticus
    May 17, 2017 at 1:31

2 Answers 2

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We have: $$ f(t) = t e^t $$ So the Laplace transform of this function is: \begin{align*} F(s) &= \mathcal{L} \{ f(t) \} \\ &= \int_0^\infty f(t) e^{-st} \, dt \\ &= \int_0^\infty t e^{t} e^{-st} \, dt \\ &= \int_0^\infty t e^{(1-s)t} \, dt \\ &= \lim_{b \to \infty} \left[ \int_0^b t e^{(1-s)t} \, dt \right] \\ &= \lim_{b \to \infty} \left(\left[ t \cdot \frac{1}{1-s}e^{(1-s)t} \right]_{t=0}^b - \int_0^b \frac{1}{1-s}e^{(1-s)t} \, dt\right) \\ &= \frac{1}{1-s}\lim_{b \to \infty} \left(b e^{(1-s)b} - \int_0^b e^{(1-s)t} \, dt\right) \\ &= \frac{1}{s-1}\lim_{b \to \infty} \int_0^b e^{(1-s)t} \, dt & (s > 1) \\ &= \frac{1}{s-1}\lim_{b \to \infty} \left[\frac{1}{1-s} e^{(1-s)t} \right]_0^b & (s > 1) \\ &= \frac{1}{(s-1)^2}\lim_{b \to \infty} \left[1-e^{(1-s)b} \right] & (s > 1) \\ &= \frac{1}{(s-1)^2} & (s > 1) \end{align*}

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To do it without integral (as in my comment in your other question), using properties of LT:

$$tf(t)\stackrel{\mathcal{L}}\longleftrightarrow-\frac{\mathrm d}{\mathrm ds}F(s)$$

Here $f(t)=e^t$ whose Laplace transform is $F(s)=\frac{1}{s-1}$.

The Laplace transform of $te^t$ becomes $$-\frac{\mathrm d}{\mathrm ds}\left(\frac{1}{s-1}\right)=\frac{1}{(s-1)^2}$$

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