convergence of integral-condition let $s \in \mathbb{C}$. We consider the integral 
$$\displaystyle\int_0^{+\infty} e^{-s x} dx$$
I read that we have 
$$\displaystyle\int_0^{+\infty} e^{-s x} dx= \dfrac{1}{s}$$
and these integral converge if $Re (s) >0$.
My question is why and how we found the condition for $s$ of convergence of this integral? Please.
 A: Intuitively, recall that if you write $e^{-sx}$ in terms of the real and imaginary parts of $s=a+bi$ you get $$e^{-(a+bi)x} = e^{-ax}e^{-ixb} = e^{-ax}(\cos(bx)-i\sin(bx)).$$ In this form, it's pretty clear from the exponential part that the integral from $x=0$  to $\infty $ converges when $a>0$ and diverges when $a<0.$ When $a=0$ the trig functions just oscillate so the integral diverges.
A: $$\int_0^\infty e^{-sx}dx=\lim_{b\to\infty}\int_0^b e^{-sx}dx=\left.\lim_{b\to\infty}-\frac1se^{-sx}\right|_0^b=$$
$$=\lim_{b\to\infty}-\frac1s\left(e^{-bs}-1\right)=\frac1s\iff \text{Re}\,s>0$$
because if we put $\;x=x+iy\;,\;\;x,y\in\Bbb R\;$ , then
$$e^{-bs}=e^{-bx-biy}\implies\left|e^{-bs}\right|=e^{-bx}\ldots$$
A: $$s=a+bi$$
$$\int_{0}^{+\infty} e^{-(a+bi)x}dx=\int_{0}^{+\infty}[e^{-ax}\cdot e^{-bxi}]dx=\int_{0}^{+\infty}[e^{-ax}\cos (-bx)+ie^{-ax}\sin (-bx)]dx=\int _{0}^{+\infty}e^{-ax}\cos (bx)dx-i\int_{0}^{+\infty}e^{-ax}\sin (bx)dx$$
$$I=\int_{0}^{+\infty}e^{-ax}\cos (bx)dx=\lim _{t\to +\infty} \int_{0}^{t}e^{-ax}\cos (bx)dx$$
$$H=\int e^{-ax}\cos (bx)dx$$
Parts:


*

*$u=e^{-ax}$

*$dv=\cos (bx)dx$


$\rightarrow$ Parts:


*

*$du=-ae^{-ax}dx$

*$v=\dfrac{1}{b}\sin (bx)$


$$H=\dfrac{1}{b}e^{-ax}\sin (bx)+\dfrac{a}{b}\int e^{-ax}\sin (bx)dx=\dfrac{1}{b}e^{-ax}\sin (bx)+\dfrac{a}{b}J$$
$$J=\int e^{-ax}\sin (bx)dx$$
Parts:


*

*$u=e^{-ax}$

*$dv=\sin (bx)dx$


$\rightarrow$ Parts:


*

*$du=-ae^{-ax}dx$

*$v=-\dfrac{1}{b}\cos (bx)$
$$J=-\dfrac{1}{b}e^{-ax}\cos (bx)-\dfrac{a}{b}\int e^{-ax}\cos (bx)dx=-\dfrac{1}{b}e^{-ax}\cos (bx)-\dfrac{a}{b}H$$
$$H=\dfrac{1}{b}e^{-ax}\sin (bx)-\dfrac{a}{b^{2}}e^{-ax}\cos (bx)-\dfrac{a^{2}}{b^{2}}H\rightarrow \dfrac{a^{2}+b^{2}}{b^{2}}H=\dfrac{1}{b}e^{-ax}\sin (bx)-\dfrac{a}{b^{2}}e^{-ax}\cos (bx)$$
$$H=\dfrac{1}{a^{2}+b^{2}}\left[ \dfrac{b\sin (bx)-a\cos (bx)}{e^{ax}}\right]\rightarrow I=\lim _{t\to +\infty} \dfrac{1}{a^{2}+b^{2}}\left[ \dfrac{b\sin (bx)-a\cos (bx)}{e^{ax}}\right]^{t}_{0}$$
$$Re(s)=a>0$$
$$I=\dfrac{1}{a^{2}+b^{2}}\left[ \lim _{t\to +\infty}\dfrac{b\sin (bt)-a\cos (bt)}{e^{at}}-(-a)\right]=\dfrac{a}{a^{2}+b^{2}}$$
Similarly: $\int_{0}^{+\infty} e^{-ax}\sin (bx)dx=\dfrac{b}{a^{2}+b^{2}}\rightarrow$
$$I=\dfrac{a}{a^{2}+b^{2}}-\dfrac{b}{a^{2}+b^{2}}i=\dfrac{1}{a+bi}=\dfrac{1}{s} (cqd)$$


For $Re(s)=a\leq 0$, the improper integrals are divergent. 
For $a=0$  is a non-existent limit and for $a<0$ the limit is $+\infty$.
