# Laplace Transform of $-e^{-at}u(-t)$

I have found a problem in applying Laplace Transform to $-e^{-at}u(-t)$ I am doing these steps:

$$= - \int_{-\infty}^{+\infty} e^{-at}u(-t) e^{-st}dt$$ $$= - \int_{-\infty}^{0} e^{-at} e^{-st}dt$$ $$= - \int_{-\infty}^{0} e^{-(a+s)t}dt$$ $$= - [-\frac{1}{a+s} e^{-(a+s)t}]|_{-\infty}^{0}$$ $$= - [-\frac{1}{a+s} (e^{-(a+s)0}-e^{-(a+s)-\infty})]$$

$$= - [-\frac{1}{a+s} (1- \infty)]$$

$$= \infty$$ Can anyone help me why it is showing like that.I check it on internet and all the books are showing the answer is $\frac{1}{s+a}$

• @Matt L can you please answer this Question? – Aadnan Farooq A Nov 25 '15 at 12:50
• That appears to be correct. The signal is anti-causal (is only non-zero for negative and zero time), but in that direction the signal is also unbounded. Can you point to an example on the internet where it gets the result you say? – Peter K. Nov 25 '15 at 13:25
• @PeterK. example You can see that on page 3, also in the book they solved this as an example – Shinning Eyes Nov 25 '15 at 13:31
• Ah! You've left out the region of convergence which is $\mathbb{R}\{s\} < -a$. If $s+a$ is negative, then the last part of your working is wrong: $e^{-(a+s)-\infty} = 0$. – Peter K. Nov 25 '15 at 13:37
• so we will check the condition for which this function can converge than will be $s+a <0$. In order to completely solve the function we should write it up to third last step and then say that it will converge for this condition.. true? – Shinning Eyes Nov 25 '15 at 13:48