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I want to evaluate the following integral via complex analysis $$\int\limits_{x=0}^{x=\infty}e^{-ax}\cos (bx)\operatorname d\!x \ \ ,\ \ a >0$$

Which function/ contour should I consider ?

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    $\begingroup$ Why not use $e^{-(a\pm ib)x}$? $\endgroup$ Feb 4 '14 at 10:14
  • $\begingroup$ And the curve ? $\endgroup$
    – WLOG
    Feb 4 '14 at 10:17
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    $\begingroup$ $[0,+\infty)$. $$\int_0^R e^{-(a\pm ib)x}\,dx = \left[-\frac{e^{-(a\pm ib)x}}{a\pm ib}\right]_0^R =\frac{1}{a\pm ib}\left(1 - e^{-(a\pm ib)R}\right) \to \frac{1}{a\pm ib}.$$ $\endgroup$ Feb 4 '14 at 10:35
  • $\begingroup$ OK... Can this be solved NOT by antidifferentiation, NOT by Laplace transform, but by contour integration? I tend to doubt it, since $x=0$ is not really a special point for the integrand. $\endgroup$
    – GEdgar
    Feb 4 '14 at 14:03
  • $\begingroup$ I've added a contour integration solution below, see if you like it! :) $\endgroup$
    – Brightsun
    Feb 17 '14 at 12:40
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As Daniel Fischer pointed out, note that

$$ \int_{0}^{+\infty} e^{-ax} \cos(bx) \; dx = \Re \Bigg( \int_{0}^{+\infty} e^{-ax} e^{ibx} \; dx \Bigg). $$

(where $\Re(z)$ denotes the real part of $z$). Then,

$$ \begin{align*} \int_{0}^{+\infty} e^{-ax} e^{ibx} \; dx & = {} \int_{0}^{+\infty} e^{-(a-ib)x} \; dx \\[1mm] & = \lim \limits_{M \to +\infty} \int_{0}^{M} e^{-(a-ib)x} \; dx \\[1mm] & = \lim \limits_{M \to +\infty} \left[ -\frac{1}{a-ib} e^{-(a-ib)x} \right]_{0}^{M} \\[1mm] & = \lim \limits_{M \to +\infty} \Big( -\frac{1}{a-ib} e^{-(a-ib)M} + \frac{1}{a-ib} \Big) \\[1mm] & = \frac{1}{a-ib} \\[1mm] & = \frac{a+ib}{\vert a-ib \vert^{2}}. \\ \end{align*} $$

So,

$$ \int_{0}^{+\infty} e^{-ax} \cos(bx) \; dx = \frac{a}{\vert a-ib \vert^{2}}. $$

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  • $\begingroup$ Can you tell me why $\int_{0}^{\infty}e^{-\alpha x}dx=\frac{1}{\alpha}$ where $\alpha$ is a complex number and $x$ is real!? $\endgroup$ Apr 12 '18 at 14:37
  • $\begingroup$ @H.R. : The function $f \, : \, t \in \mathbb{R} \, \mapsto \, e^{-\alpha t}$ is defined on the real line. Its derivative exists everywhere and: $\forall t \in \mathbb{R}, \; f'(t) = - \alpha e^{-\alpha t}$. Then, if $\alpha \neq 0$, $$ t \in \mathbb{R} \; \mapsto \; - \frac{1}{\alpha} e^{-\alpha t} $$ is an antiderivative of $f$. Eventually: $$ \int_{0}^{+\infty} e^{-\alpha t} \; dt = \lim \limits_{M \to +\infty} \left[ - \frac{1}{\alpha} e^{-\alpha t} \right]_{t=0}^{t=M} = \lim \limits_{M \to +\infty} -\frac{1}{\alpha} e^{-\alpha M} + \frac{1}{\alpha} = \frac{1}{\alpha}. $$ $\endgroup$
    – pitchounet
    Apr 13 '18 at 15:43
  • $\begingroup$ I don't see why $\lim_{M\to+\infty}e^{-\alpha M}=0$? Please note that $\alpha$ is a complex number not a real one. I understand the identity when $\alpha$ is real but have no idea for the complex case. $\endgroup$ Apr 13 '18 at 19:36
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    $\begingroup$ @H.R. : If $\alpha \in \mathbb{C}^{\ast}$ and $M \in \mathbb{R}^{+}$, $$ \left| e^{-\alpha M} \right| = e^{- \Re(\alpha) M }. $$ Therefore, as long as $\Re(\alpha) > 0$, you get the desired result. Note that the condition $\Re(\alpha) > 0$ ensures that the integral $$ \int_{0}^{+\infty} e^{-\alpha x} \; dx $$ is convergent. $\endgroup$
    – pitchounet
    Apr 14 '18 at 4:07
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Let us integrate the function $e^{-Az}$, where $A=\sqrt{a^2+b^2}$ on a circular sector in the first quadrant, centered at the origin and of radius $\mathcal{R}$, with angle $\omega$ which satisfies $\cos \omega = a/A$, and therefore $\sin \omega = b/A$. Let this sector be called $\gamma$.

Since our integrand is obviously holomorphic on the whole plane we get: $$ \oint_\gamma \mathrm{d}z e^{-Az} = 0. $$ Breaking it into its three pieces we obtain: $$ \int_0^\mathcal{R}\mathrm{d}x e^{-Ax}+\int_0^\omega \mathrm{d}\varphi i\mathcal{R}e^{i\varphi}e^{-A\mathcal{R}e^{i\varphi}}+\int_{\mathcal{R}}^0 \mathrm{d}r e^{i\omega}e^{-Are^{i\omega}}=0. $$ The mid integral, as $\mathcal{R}\to\infty$ is negligible. So: $$ \int_0^\infty\mathrm{d}xe^{-Ax}=\int_0^\infty\mathrm{d}r (\cos\omega+i\sin\omega)e^{-Ar(\cos\omega+i\sin\omega)} $$ $$ \frac{1}{A}=\frac{1}{A}\int_0^\infty\mathrm{d}r(a+ib)e^{-r(a+ib)} $$ $$ \int_0^\infty\mathrm{d}r(a+ib)e^{-ar} (\cos br - i\sin br) = 1 $$ Now let's call $I_c = \int_0^\infty\mathrm{d}re^{-ar}\cos br$ and $I_s = \int_0^\infty\mathrm{d}re^{-ar}\sin br$, then: $$ aI_c-iaI_s+ibI_c+bI_s=1 $$ and by solving: $$ aI_c+bI_s=1;\ \ \ \ -aI_s+bI_c=0 $$ $$ I_c=\frac{a}{a^2+b^2}; \ \ \ \ I_s=\frac{b}{a^2+b^2}. $$ This method relies only on the resource of contour integration as you asked!

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  • $\begingroup$ I have come to this problem when I am doing the exercises on Stein's complex analysis. And using this hint, I could be able to calculate the integral. However, in general, how can I think of integrating $e^{-Az}$ just from the original problem? Or put it this way, are there any deep reason why we should think of this particular function? $\endgroup$
    – Jason
    Nov 15 '17 at 6:14
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First,you may use the definition of Lapace transform to get it or integration by parts twice.
For complex numbers your integrand is the real part of $\exp(-ax+ibx)$. Use this function to evaluate the unbounded integral then evaluate the bounded one

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