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What would $\int\limits_{-\infty}^\infty e^{ikx}dx$ be equal to where $i$ refers to imaginary unit? What steps should I go over to solve this integral?

I saw this in the Fourier transform, and am unsure how to solve this.

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  • $\begingroup$ As an indefinite integral: can you integrate $f(ax)$ in terms of the integral of $f(x)$? You need substitution. It works just as fine with $a$ complex. As an integral over $\Bbb R$: you need distribution theory to make sense of the Fourier transform of $1$. $\endgroup$
    – anon
    Jul 31, 2012 at 8:52
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    $\begingroup$ are you missing integration limits? For $\int_{-\infty}^\infty e^{ikx}dx$ see Dirac's $\delta$ function. Otherwise this is quite easy...BTW: Welcome to Math.StackExchange.com $\endgroup$
    – draks ...
    Jul 31, 2012 at 8:53

7 Answers 7

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$$ \begin{align} \int_{-\infty}^\infty e^{ixy}\overbrace{\ \ \ e^{-\epsilon x^2}\ \ \ }^{\to1}\,\mathrm{d}x &=e^{-\frac{y^2}{4\epsilon}}\color{#C00}{\int_{-\infty}^\infty e^{-\epsilon\left(x-\frac{iy}{2\epsilon}\right)^2}\,\mathrm{d}x}\tag{1}\\ &=\underbrace{\color{#C00}{\sqrt{\frac\pi\epsilon}}\,e^{-\frac{y^2}{4\epsilon}}}_{\to2\pi\delta(y)}\tag{2} \end{align} $$ Using Cauchy's Integral Theorem, the red integral in $(1)$ is simply $\int_{-\infty}^\infty e^{-\epsilon x^2}\,\mathrm{d}x=\sqrt{\frac\pi\epsilon}$ .

As $\epsilon\to0$, we get that $(2)$ approximates $2\pi\delta(y)$. That is, the integral of $(2)$ is $2\pi$ for all $\epsilon$, and as $\epsilon\to0$, the main mass of the function is squeezed into a very small region about $0$.

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Draks is right on this (+1) it makes sense as (up to a constant) a representation of the Dirac delta distribution (it is divergent from other points of view!).

More exactly : $$2\pi \delta(k)=\int\limits_{-\infty}^\infty e^{ikx}dx$$

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  • $\begingroup$ What if we change the limits as zero to infinity? $\endgroup$ Jan 3, 2017 at 6:23
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    $\begingroup$ @ZohaibAarfi: in this case you may for example replace the integral by $$I_{\epsilon}(k):=\int_{0}^{\infty}e^{(-\epsilon+ik) x}\,dx$$ and take the limit as $\epsilon\to 0$ ($\epsilon>0$) as done here (with $t:=-x$) or in this thread (with $w=-k$) or this answer. Hoping this helped, $\endgroup$ Jan 3, 2017 at 8:52
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$\int\limits_{-\infty}^\infty e^{ikx}dx$ is a (EDIT: scaled by $\frac1{2\pi}$) representation of Dirac's $\delta(k)$ function. For the antiderivative see the other answers...

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I just want to clarify what is meant by saying that "as a distribution", $$ 2\pi\delta(k) = \int_{-\infty}^\infty e^{ikx}\,dx. $$ The right hand side certainly doesn't make sense, and the left hand side can only make sense for $k\not=0$. What we mean though is that for any smooth function $\phi$ with rapid decay, $$ 2\pi \phi(0) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{ikx} \phi(k) \,dk\,dx. $$ Now both sides make sense. This isn't trivial to prove, and in fact it implies the Fourier inversion formula (using the relationship between modulation and translation under the Fourier transform). The proof often goes through a calculation similar to what @robjohn's answer describes.

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  • $\begingroup$ For $\phi, \phi' \in L^1$ there is a one line proof. Convolution by a mollifier $ne^{-\pi n^2 x^2}$ works well if $\phi,\widehat{\phi} \in L^1$ $\endgroup$
    – reuns
    Sep 5, 2017 at 21:05
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The definition of the Fourier inverse:

$$ f(t) = \mathfrak{F}^{-1}\{F(jw)\} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(jw) e^{jwt} dw $$

The Fourier pair (in the angular frequency domain):

$$ \delta(t) \leftrightarrow 1 $$

The integral in the question:

$$ 2\pi \times \frac{1}{2\pi} \int_{-\infty}^{+\infty} 1 \times e^{jxk} dx = 2\pi \times \delta(k) = 2\pi \delta(k)$$

The variable substitution $k=t$ was made and the u-substitution $w=x$ was made, for clarity.

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Formally it is solved exactly as in the real case. The antiderivative is $\frac{1}{i k} e^{ikx}$. So, $\int e^{ikx} dx = \frac{1}{i k} e^{ikx} + C$.

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  • $\begingroup$ To the downvoter, the OP changed the question after I answered it. If you need to downvote, at least have the whatever to explain your actions. Actions like this make me want to stop helping. $\endgroup$
    – copper.hat
    Feb 14, 2020 at 6:46
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Isn't it

$$ -{i\over k} e^{i k x} + C $$

Just start by deriving

$$ e^{i k x} $$

And workout what constants are needed to adjust the result or alternatively use the formula by noticing that you are integrating

$$ e^{c x} $$

with:

$$ c=ik $$

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