# Inverse Laplace Transform of $1/(s+1)$ without table

The pole is on the left half plane, so $\gamma =0$

$$\frac{1}{2i\pi}\int ^{i\infty}_{-i\infty}\frac {e^{st}}{s+1}ds$$

substituting $iu=s$

$$\frac{1}{2i\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}idu$$

(EDIT: I managed to get the answer $e^{-t}$ using feynman technique, but I dropped some conditional somewhere I guess.)I am unable to integrate this, however mathematica gives

$$\frac12 e^{-\left| t\right| } (\text{sgn}(t)+1)$$

which is correct for $t>0$ How do I get this result?

I also have a more general request:

I was unable to find a site which shows how to perform these inverse laplace transforms of simple functions, like $\frac{1}{s-a}$. When I search for inverse laplace transform, I either see the formula for it (which isn't all that clear to me right now) or a table. I would like to learn to how to do these transforms.

• I am afraid I am not following what it is that complex numbers are doing here... For a direct solution, one might be aware that, for every positive real number $a$ (or, for every complex number $a$ with $\Re(a)>0$...), $$\int_0^\infty e^{-at}dt=\frac1a,$$ hence $$\frac1{s+1}=\int_0^\infty e^{-(s+1)t}dt=\int_0^\infty e^{-st}e^{-t}dt,$$ which shows the solution is $$t\mapsto e^{-t}\mathbf 1_{t>0}.$$ Or you are using another definition of the Laplace transform, and then this should be explained in the question.
– Did
Commented May 18, 2015 at 7:15
• You forward transformed the answer, which is quite a bit easier than inverse transforming $1/(s+1)$ Commented May 18, 2015 at 8:18
• No idea what "forward transform" means exactly but, yes, to keep constantly in one's mind that $$\frac1a=\int_0^\infty e^{-at}dt$$ is something I would try to achieve if I were interested in Laplace transforms.
– Did
Commented May 18, 2015 at 9:08
• "The pole is on the left half plane" - what does this mean and how do you know this? thank you so much! Commented Jul 14, 2022 at 5:35

Let continue your calculus : $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}du$$ $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {(\cos(ut)+i \sin(ut))(-iu+1)}{u^2+1}du$$ $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {\cos(ut)+u \sin(ut)}{u^2+1}du + \frac{i}{2\pi}\int ^{\infty}_{-\infty}\frac {-u\cos(ut)+ \sin(ut)}{u^2+1}du$$

$\int ^{\infty}_{\infty}\frac {-u\cos(ut)}{u^2+1}du =0$ (integral of odd function)

$\int ^{\infty}_{-\infty}\frac {\sin(ut)}{u^2+1}du =0$ (integral of odd function)

Then, with the known integrals :

$\int ^{\infty}_{-\infty}\frac {\cos(ut)}{u^2+1}du=2\int ^{\infty}_0\frac {\cos(ut)}{u^2+1}du=2\left(\frac{\pi}{2}e^{-t}\right)$ in $t>0$

$\int ^{\infty}_{-\infty}\frac {u \sin(ut)}{u^2+1}du=2\int ^{\infty}_0\frac {u \sin(ut)}{u^2+1}du=2\left(\frac{\pi}{2}e^{-t}\right)$ in $t>0$

Finally, in $t>0$ : $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}du=\frac{1}{2\pi}\left( 2\left(\frac{\pi}{2}e^{-t}\right) + 2\left(\frac{\pi}{2}e^{-t}\right) \right) = e^{-t}$$ .

Note : in my first edition, there was a mismatch of notations : In fact $y=t$.

• Lets say I want to verify: $\int ^{\infty}_{-\infty}\frac {\cos(ut)}{u^2+1}du= \pi e^{-y}$ in $y>0$. I pick the upper half circle contour, and consider $\cos(ut)=e^{iut}$ since the integral will be symmetrical. The residue at u=i will be $\frac{e^{-t}}{2i}$, multiplying this by $2\pi i$ I get $\pi e^{-t}$. Where did the $e^{-|t|}$ go, I lose track of the conditional. Commented May 18, 2015 at 8:57
• Since the function is even, In the case $t<0$, ether use $\cos(ut)=\cos(-ut)=e^{-iut}$, or change of variable $t=-\tau$ and proceed on the same manner, which leads to $e^t$ instead of $e^{-t}$. So, both together $e^{-|t|}$ Commented May 18, 2015 at 9:33