The Laplace transform of the integral of f(t) is given by the following formula L{∫_{0}^{t }f(u) du} = F(s)/s. Here we prove this Laplace formula of integrals.

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## Statement

The Laplace transform of an integral is given by

$L\Big\{ \int_0^t f(u) du \Big\} = \dfrac{F(s)}{s}$

where L{f(t)} = F(s) denotes the Laplace transform of f(t).

## Proof

Let us put

g(t) = $\int_0^t f(u) du$.

Note that g(0)=0. By the fundamental theorem of Calculus, we have that

g$’$(t)=f(t).

Now, using the Laplace Transform of derivatives, we have that

L{g$’$(t)} = -g(0) + s L{g(t)}

⇒ L{f(t)} = -0 + s L{g(t)} as we have g(0)=0 and g$’$(t)=f(t).

⇒ F(s) = s L{g(t)}

⇒ L{g(t)} = $\dfrac{F(s)}{s}$

This proves that L{∫_{0}^{t }f(u) du} = F(s)/s and this is the Laplace transform of integrals.

Have You Read These?

Laplace Transform: Definition, Table, Formulas, Properties

Solved problems of Laplace transforms

Laplace Transform of derivatives

## FAQs

**Q1: What is the Laplace transform of integral formula?**

Answer: If L{f(t)} = F(s), then the Laplace of the integral of f(t) is given by L{∫_{0}^{t }f(u) du} = F(s)/s.