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I am having trouble solving this integral with complex analysis

$$\int\limits^{\infty }_{0}\frac{e^{iax}}{x^{2} +1} dx$$

I have tried two different contours; those being

contour 2

contour 1

with both contours, I got the answer $\int\limits ^{\infty }_{0}\frac{e^{iax}}{x^{2} +1} dx=\frac{\pi }{2e^{a}}$ But according to wolfram alpha's approximation, this is wrong. Wolfram alpha had a real and imaginary part to the answer. So I'm confused. Can someone show the process for solving this with contour integration?

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  • $\begingroup$ What is $a$? Any complex number? $\endgroup$ – Mark Apr 16 at 18:41
  • $\begingroup$ a is a real number or else it would diverge $\endgroup$ – Highvoltagemath Apr 16 at 18:42
  • $\begingroup$ Note that $\Im\int_0^\infty\frac{\exp(ix)}{x^2+1}\,dx=\int_0^\infty\frac{\sin x}{x^2+1}\,dx>0$ so the integral cannot be purely real. $\endgroup$ – TheSimpliFire Apr 16 at 18:54
  • $\begingroup$ Sorry about the missing $a$ term in the title. $\endgroup$ – Highvoltagemath Apr 16 at 18:56
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In contour 1, I am guessing you treated the integral along the segment $C1$ as zero, but it is not, and it has a non-zero imaginary part.

Similarly, the same integral, with imaginary result, is omitted along $C1$ in contour 2.

The integral does not have a real value.

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  • $\begingroup$ But can it still be solved by contour integration? $\endgroup$ – Highvoltagemath Apr 16 at 19:00
  • $\begingroup$ Do you know how to solve it with complex analysis? $\endgroup$ – Highvoltagemath Apr 16 at 20:13
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The exponential integral $E_1(z):=\int_z^\infty\frac{e^{-t} dt}{t}=e^{-z}U(1,\,1,\,z)$ in terms of a confluent hypergeometric function of the second kind. Since $\frac{1}{x^2+1}=\frac{i}{2}\sum_\pm\frac{\pm 1}{x\pm i}$, your integral is $$\frac{i}{2}\sum_\pm\pm\int_0^\infty\frac{e^{iax}dx}{x\pm i}=\frac{i}{2}\sum_\pm\pm e^{\pm a}\int_{\pm a}^\infty\frac{e^{-z}dz}{z}\\=\frac{i}{2}\sum_\pm\pm e^{\pm a}E_1(\pm a)=\frac{i}{2}\left(U(1,\,1,\,a)-U(1,\,1,\,-a)\right).$$This isn't real, but its real part is $-\frac{\pi}{2}e^{-a}$ for $a>0$. This means I probably have a sign error somewhere editors are welcome to address, since $$\int_0^\infty\frac{\cos axdx}{x^2+1}=\frac12\Re\int_{\Bbb R}\frac{e^{iax}dx}{x^2+1}=\frac{\pi}{2}e^{-|a|}$$is a useful sanity check based on the Cauchy distribution's characteristic function.

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  • $\begingroup$ so that means we get the imaginary part by solving $\int\limits ^{\infty }_{0}\frac{i\cdot sin( x)}{x^{2} +1} dx$, one problem is that I don't see how to solve this part. $\endgroup$ – Highvoltagemath Apr 16 at 21:54
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I can't directly use contour integration to solve this, but I can use Fourier Transforms and their properties.

$$\begin{align*}I(a) &= \int_0^\infty \dfrac{e^{iax}}{x^2+1} dx \\ \\ &= 2\pi \int_0^\infty \dfrac{e^{i2\pi sa}}{\left(2\pi s\right)^2 +1} ds \\ \\ &= \pi \int_{-\infty}^\infty H(s)\dfrac{2}{\left(2\pi s\right)^2 +1} e^{i2\pi sa}ds \\ \\ &= \pi \mathscr{F}^{-1}\left\{H(s)\dfrac{2}{\left(2\pi s\right)^2 +1}\right\}\\ \\ &= \pi \cdot \mathscr{F}^{-1}\left\{H(s)\right\}*\mathscr{F}^{-1}\left\{\dfrac{2}{\left(2\pi s\right)^2 +1}\right\}\\ \\ &= \pi \cdot \dfrac{1}{2}\left[\dfrac{1}{i\pi(-a)} + \delta(-a)\right] * e^{-|a|}\\ \\ &= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \dfrac{1}{a} * e^{-|a|}\\ \\ &= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \int_{-\infty}^\infty \dfrac{e^{-|\tau|}}{a-\tau} d\tau\\ \\ &= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2} \int_0^\infty \dfrac{e^{-\tau}}{a-\tau} d\tau+ \dfrac{i}{2} \int_{-\infty}^0 \dfrac{e^{\tau}}{a-\tau} d\tau\\ \\ &= \dfrac{\pi}{2} e^{-|a|} - \dfrac{i}{2} \int_{-a}^\infty \dfrac{e^{-(t+a)}}{t} dt - \dfrac{i}{2} \int_{\infty}^a \dfrac{e^{(a-t)}}{t} dt\\ \\ &= \dfrac{\pi}{2} e^{-|a|} - \dfrac{i}{2}e^{-a} \int_{-a}^\infty \dfrac{e^{-t}}{t} dt + \dfrac{i}{2} e^{a}\int_a^{\infty} \dfrac{e^{-t}}{t} dt\\ \\ &= \dfrac{\pi}{2} e^{-|a|} + \dfrac{i}{2}e^{-a} \mathrm{Ei}(a) - \dfrac{i}{2} e^{a}\mathrm{Ei}(-a)\\ \\ &= \dfrac{\pi}{2} e^{-|a|} -i \space \dfrac{e^{a} \mathrm{Ei}(-a) - e^{-a}\mathrm{Ei}(a)}{2}\\ \end{align*}$$

This answer looks like what @J.G. was driving at.

Note that just deriving the inverse Fourier Transform of the Heaviside unit step, $H(s)$, involves multiple contour integrations and distribution theory.

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  • $\begingroup$ Thank you for solving this! But I do have one question: why with the two contours I tried, could only get the real part of the integral? $\endgroup$ – Highvoltagemath Apr 16 at 23:50
  • $\begingroup$ You need to be able to compute and consider all the other segments that make up the closed contour. If you cannot, then you can't get an expression for the segment for which you are interested. $\endgroup$ – Andy Walls Apr 17 at 0:09
  • $\begingroup$ The problem is that the natural contour would involve the whole real axis (coming back around at infinity in the upper half plane. This gives you $\int_{-\infty}^0 f(x) dx + \int_0^{\infty} f(x) dx$. The first of those is easily seen to be the complex conjugate of the second. So the contour integral you can get gives you the integral you want, plus its complex conjugate. This makes it easy to find the real part (which you did) but says nothing about the imaginary part. $\endgroup$ – Mark Fischler Apr 27 at 5:33
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Note that} \\[2mm] &\ \left.\int_{0}^{\infty}{\expo{\ic ax} \over x^{2} + 1}\,\dd x \,\right\vert_{\ a\ \in\ \mathbb{R}} \\[2mm] = &\ \underbrace{\int_{0}^{\infty}{\cos\pars{\verts{a}x} \over x^{2} + 1}\,\dd x}_{\ds{\pi\expo{-\verts{a}} \over 2}}\ +\ \ic\,\mrm{sgn}\pars{a} \bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x}\label{1}\tag{1} \end{align}


Then, \begin{align} &\bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x} = \overbrace{-\Im\lim_{R \to \infty}\int_{0}^{\pi/2}{\exp\pars{\ic\verts{a}R\expo{\ic\theta}} \over R^{2}\expo{2\ic\theta} + 1}\,R\expo{\ic\theta}\ic\,\dd\theta} ^{\ds{=\ 0}} \\[2mm] &\ -\Im\lim_{\epsilon \to 0^{+}}\bracks{% \int_{\infty}^{1 + \epsilon}{\expo{-\verts{a}y} \over -y^{2} + 1}\,\ic\,\dd y + \int_{1 - \epsilon}^{0}{\expo{-\verts{a}y} \over -y^{2} + 1}\,\ic\,\dd y} \\[8mm] = &\ -\mrm{P.V.}\int_{0}^{\infty}{\expo{-\verts{a}y} \over y^{2} - 1} \,\dd y = -\,{1 \over 2}\,\mrm{P.V.}\int_{0}^{\infty} {\expo{-\verts{a}y} \over y - 1}\,\dd y + {1 \over 2}\int_{0}^{\infty} {\expo{-\verts{a}y} \over y + 1}\,\dd y \\[5mm] = &\ -\,{1 \over 2}\,\expo{-\verts{a}}\mrm{P.V.}\int_{-\verts{a}}^{\infty} {\expo{-y} \over y}\,\dd y + {1 \over 2}\,\expo{\verts{a}}\ \underbrace{\int_{\verts{a}}^{\infty} {\expo{-y} \over y}\,\dd y}_{\ds{\mrm{E}_{1}\pars{\verts{a}}}} \end{align}

$\ds{\mrm{E}_{1}}$ is the Exponential Integral.

Then, \begin{align} &\bbox[10px,#ffd]{\Im\int_{0}^{\infty}{\expo{\ic\verts{a}x} \over x^{2} + 1}\,\dd x} \\[5mm] = &\ -\,{1 \over 2}\,\expo{-\verts{a}}\bracks{% \mrm{P.V.}\int_{-\verts{a}}^{\verts{a}}{\expo{-y} \over y}\,\dd y + \mrm{E}_{1}\pars{\verts{a}}} + {1 \over 2}\,\expo{\verts{a}}\ \mrm{E}_{1}\pars{\verts{a}} \\[5mm] = &\ -\,{1 \over 2}\,\expo{-\verts{a}} \int_{0}^{\verts{a}}\pars{{\expo{-y} \over y} + {\expo{y} \over -y}} \,\dd y + \sinh\pars{\verts{a}}\mrm{E}_{1}\pars{\verts{a}} \\[5mm] = &\ \expo{-\verts{a}} \int_{0}^{\verts{a}}{\sinh\pars{y} \over y}\,\dd y + \sinh\pars{\verts{a}}\,\mrm{E}_{1}\pars{\verts{a}} \\[5mm] = &\ \expo{-\verts{a}}\,\mrm{Shi}\pars{\verts{a}} + \sinh\pars{\verts{a}}\,\mrm{E}_{1}\pars{\verts{a}} \label{2}\tag{2} \end{align}

$\ds{\mrm{Shi}}$ is the Hyperbolic Sine Integral.


\eqref{1} and \eqref{2} lead to: $$ \bbx{\left.\int_{0}^{\infty}{\expo{\ic ax} \over x^{2} + 1}\,\dd x \,\right\vert_{\ a\ \in\ \mathbb{R}} = {\pi\expo{-\verts{a}} \over 2} + \bracks{\vphantom{\Large A}\expo{-\verts{a}}\,\mrm{Shi}\pars{a} + \sinh\pars{a}\,\mrm{E}_{1}\pars{\verts{a}}}\ic} $$

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  • $\begingroup$ What is P.V.? and also, it seems you used a contour with 3 parts right? but what was your contour? $\endgroup$ – Highvoltagemath Apr 17 at 11:56
  • $\begingroup$ @Highvoltagemath Cauchy principal value. $\endgroup$ – J.G. Apr 17 at 18:27

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