The Laplace transform of 1/t does not exist. In this post, using the gamma function we will prove that the Laplace of 1 over t does not exist.
Table of Contents
Laplace of 1/t
| Question: Show that the Laplace transform of 1/t does not exist. |
Answer:
By definition of Laplace transforms, the Laplace of 1/t will be equal to
L{1/t} = $\int_0^\infty \dfrac{1}{t}$ e-stdt …(I)
Put u=st.
So du=s dt ⇒ dt = du/s.
Also, t=u/s.
| t | u |
| 0 | 0 |
| ∞ | ∞ |
So from (I) we have that
L{1/t} = $\int_0^\infty \dfrac{s}{u} e^{-u} \dfrac{du}{s}$
= $\int_0^\infty u^{-1} e^{-u} du$
= $\int_0^\infty u^{0-1} e^{-u} du$
= Γ(0) as the Gamma function is defined by Γ(z) = $\int_0^\infty$ tz-1 e-t dz. So we have obtained that
| L{1/t} = Γ(0). |
Final Conclusion:
We know Γ(n) = (n-1)!, so Γ(0) is not defined.
As L{1/t}=Γ(0), we conclude that the Laplace transform of 1/t is not defined, hence the Laplace of 1/t does not exist.
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FAQs
Answer: The Laplace transform of 1/t does not exist. This is because the Laplace of 1/t can be expressed in terms of the gamma function Γ(-1) which does not exist.
Answer: L{1/t} does not exist.
