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 $\frac{1}{s+a}$
Usually Laplace transform is defined as $$\mathcal{L} f= \int_{0}^{+\infty} f(t) e^{-st}dt $$ The transform $${\mathcal{B}} f= \int_{-\infty}^{+\infty} f(t) e^{-st}dt $$ is a bilateral Laplace transform.
For a Laplace transform $\mathcal{L} e^{at}f = F(s-a)$ where $F=\mathcal{L} f$, i.e. a frequency shifting. Similarly $\mathcal{L} e^{-at}f = F(s+a)$
Now, you talking about $u(-t)$. As I could guess from your integration and since this is a pretty common notation for a Laplace transform discussion, $u(t)$ is a step function, also called Heaviside step function.
Now, $\mathcal{L} u(t)=\frac{1}{s}$, therefore $\mathcal{L} (e^{-at}u(t))=\frac{1}{s+a}$. However, $\mathcal{L} u(-t) = 0$.
Note
In your integration you get (note that you assume $s+a\ne 0$. The $s+a=0$ is a special case which should give $\infty$)
$$\frac{1}{a+s}e^{-(a+s)t} \Bigg|_{-\infty}^0 =\frac{1}{a+s}\left(e^0-\lim_{r\to\infty} e^{(a+s)r}\right) =\begin{cases} \infty &a+s>0\\ \frac{1}{a+s}&a+s<0 \end{cases} $$ since $$\lim_{r\to\infty} e^{-(a+s)r} = \begin{cases} \infty &a+s>0\\ 0&a+s<0 \end{cases}$$