# Laplace transform of $\frac{1}{C}\int_0^1i(t)dt$

I have the integral:

$$\frac{1}{C}\int_0^1i(t)dt$$ which I should transform with Laplace. There is a rule saying that $$\int_0^ti(t)dt$$ has the transform $$s^{-1}F(s)$$ can I use this to transform my integral? The only difference is that t is 1, and I dont really know how that affect the rule stated above.

Note that the Laplace transform, transforms a function from the $t$-domain to the $s$-domain, now, notice that $\int\limits_0^1 {i(t)dt}$ is not a varying function of $t$, but a constant. Hence, its Laplace transform is equal to $\frac{1}{s}$ multiplied by itself (I suggest you use different notations for the dummy variable of the integral to avoid such ambiguities).
• I think there should be a $1/s$ in front of it. – Urgje Oct 5 '14 at 11:22
• @Urgje You are right, since the Laplace of a constant $a$ is $\frac{a}{s}$. – Seyed Mohsen Ayyoubzadeh Oct 5 '14 at 11:25
• So, is this the correct answer: $L\{\int_0^1 i(t)dt\} = \frac{1}{s}\int_0^1 I(s)dt$? – theva Oct 5 '14 at 12:37
• @theva The suggestion I made was precisely for not making this mistake. Note that $F(s) = L\{ f(t)\} = \int\limits_0^{ + \infty } {f(t){e^{ - st}}dt}$, so that here $$L\{ f(t)\} = \int\limits_0^{ + \infty } {\left( {\frac{1}{C}\int\limits_0^1 {i(\tau )d\tau } } \right){e^{ - st}}dt} = \left( {\frac{1}{C}\int\limits_0^1 {i(\tau )d\tau } } \right)\left( {\int\limits_0^{ + \infty } {{e^{ - st}}dt} } \right)$$Hope it helps ;) [Again note that ${\int\limits_0^1 {i(\tau )d\tau } }$ has nothing to do with $I(s)$ which is $I(s) = \int\limits_0^{ + \infty } {i(t){e^{ - st}}dt}$] – Seyed Mohsen Ayyoubzadeh Oct 5 '14 at 13:44