y(t) a of an integrator circuit I have a signal $$ x(t)= \frac{1}{T} e^{-\frac{t}{T}} u(t) - \frac{1}{T} e^{\frac{t}{T}} u(-t) $$ 

and I know that it transits in a integrator circuit and I have to find y(t) in time and frequency domain. From theory on my book i know that 

$$ y(t) = \int_{-\infty}^{t} x(\tau) d \tau $$

So I started to calculate this integral with my $x(t)$:

$$ \frac{1}{-RCT} \left[ \int_{0}^{+\infty} e^{\frac{-\tau}{T}} d\tau - \int_{—\infty}^{0} e^{\frac{\tau}{T}}d\tau \right] $$ 

but this gave me 

$$ \frac{1}{-RCT} \left[  \dfrac{e^{-\infty } -1}{-\frac{1}{T}} - \dfrac{-e^{-\infty } +1}{\frac{1}{T}} \right] $$ 

and this gave me 0. 

The correct result should be $$ -e^{ -\frac{|t|}{T} } $$

Thank you so much
The input and the integrator output are shown below:



So, you should consider two cases: $t&lt;0$ and $t&gt;0$. 

In the first case ($t&lt;0$), you have 
$$\begin{align}
y(t) &amp;= \int_{-\infty}^t x(\tau)d\tau = -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}u(-\tau)d\tau \\
&amp;= -\frac{1}{T}\int_{-\infty}^t e^{\frac{\tau}{T}}d\tau = -\frac{1}{T}Te^{\frac{\tau}{T}}\Big]_{-\infty}^t \\
&amp;= -e^{\frac{\tau}{T}}\Big]_{-\infty}^{t} = -(e^{\frac{t}{T}} - 0) = -e^{\frac{t}{T}}
\end{align}$$
again, for $t&lt;0$. So now you have
$$y(t) = \left\{\begin{array}{ll}
-e^{\frac{t}{T}}, &amp; t &lt; 0 \\
? \quad \:\:\:, &amp; t &gt; 0
\end{array}\right.$$
and you're looking for the other branch of the output. Can you show that it is $$y(t) = -e^{-\frac{t}{T}}, \quad t &gt; 0?$$

If you do, then you'll get
$$y(t) = \left\{\begin{array}{ll}
-e^{\frac{t}{T}}, &amp; t &lt; 0 \\
-e^{-\frac{t}{T}},  &amp; t &gt; 0
\end{array}\right.= -e^{-\frac{|t|}{T}}$$ for all $t$.