I'm currently trying to find the inverse Laplace transform for:

$$\frac{1}{s + 1}e^{-s}$$

The answer says that the inverse Laplace transform is:

$$\mathscr{L}^{-1}\left( \frac{1}{s + 1}e^{-s} \right) = e^{t - 1}u(t - 1)$$

I'm aware that the Heaviside function's transform is:

$$ \begin{align} \mathscr{L}(u(t - a)) & = \frac{1}{s}e^{-as} \\ \mathscr{L}(f(t - a) u(t - a)) & = e^{-as}F(s) \end{align} $$

but I'm having trouble figuring out how the inverse transform was derived. Any tips are appreciated. Thanks in advance!

  • $\begingroup$ Observe that $a=1$ because the transform is $e^{-as}F(s)$. $\endgroup$
    – Axion004
    Commented Dec 20, 2019 at 3:20
  • 1
    $\begingroup$ The answer as you've reported it is incorrect: The Laplace transform of $e^{t-1}u(t-1)$ is $e^{-s}/(s-1)$, not $e^{s}/(s+1)$. $\endgroup$ Commented Dec 20, 2019 at 3:23
  • 2
    $\begingroup$ As a tipoff to the answer being wrong, note that your Laplace transform diverges at $s=-1$ but is convergent for $s>1$. That means that the integral $\int_0^\infty e^{t}f(t)\,dt$ is divergent, but also that the integral would be finite if the exponent were any smaller. This is obviously incompatible with $f(t)$ growing like $e^{t}$, but it is entirely compatible with $f(t)$ decaying like $e^{-t}$. $\endgroup$ Commented Dec 20, 2019 at 3:31

1 Answer 1


Hint: Apply the transform in reverse

$$\mathcal{L}^{-1}\left( e^{-as}F(s) \right) = f(t-a)u(t-a)$$

where $$F(s)=\frac{1}{s+1}$$


$$f(t)=\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}\left( \frac{1}{s+1}\right)=e^{-t}$$

Can you finish?


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