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^{-st}dt …(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) = $\int_0^\infty$ t^{z-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

**Q1: Does the Laplace transform of 1/t exist?**

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.

**Q2: What is L{1/t}?**

Answer: L{1/t} does not exist.