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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.

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  • $\begingroup$ Sorry but i don't understand your idea $\endgroup$ – user415040 Apr 1 '17 at 17:13
  • $\begingroup$ I assume that $x$ is real. Can you anti derive that e-power function? That's just a start.... $\endgroup$ – imranfat Apr 1 '17 at 17:14
  • $\begingroup$ yes, $x$ is real. We have $(e^{sx})'= s e^{sx}$ $\endgroup$ – user415040 Apr 1 '17 at 17:17
  • $\begingroup$ But you took a derivative? $\endgroup$ – imranfat Apr 1 '17 at 17:17
  • $\begingroup$ space is dark addressed the second part as to why RE(s) should be positive. I took out my first comment... $\endgroup$ – imranfat Apr 1 '17 at 17:23
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$$s=a+bi$$ $$\int_{0}^{+\infty} e^{-(a+bi)x}dx=\int_{0}^{+\infty}[e^{-ax}\cdot e^{-bxi}]dx=\int_{0}^{+\infty}[e^{-ax}\cos (-bx)+ie^{-ax}\sin (-bx)]dx=\int _{0}^{+\infty}e^{-ax}\cos (bx)dx-i\int_{0}^{+\infty}e^{-ax}\sin (bx)dx$$ $$I=\int_{0}^{+\infty}e^{-ax}\cos (bx)dx=\lim _{t\to +\infty} \int_{0}^{t}e^{-ax}\cos (bx)dx$$ $$H=\int e^{-ax}\cos (bx)dx$$

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 $+\infty$.

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  • $\begingroup$ I think it is fine but a little too lengthy. $\endgroup$ – DonAntonio Apr 1 '17 at 20:58
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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.

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  • $\begingroup$ Can you look and comment my answer please? $\endgroup$ – Guillemus Callelus Apr 1 '17 at 18:25
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$$\int_0^\infty e^{-sx}dx=\lim_{b\to\infty}\int_0^b e^{-sx}dx=\left.\lim_{b\to\infty}-\frac1se^{-sx}\right|_0^b=$$

$$=\lim_{b\to\infty}-\frac1s\left(e^{-bs}-1\right)=\frac1s\iff \text{Re}\,s>0$$

because if we put $\;x=x+iy\;,\;\;x,y\in\Bbb R\;$ , then

$$e^{-bs}=e^{-bx-biy}\implies\left|e^{-bs}\right|=e^{-bx}\ldots$$

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  • $\begingroup$ Can you look and comment my answer please? $\endgroup$ – Guillemus Callelus Apr 1 '17 at 18:25

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