10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '12 at 8:52
  • 2
    $\begingroup$ 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 '12 at 8:53
  • $\begingroup$ @draks oh.. forgot to write integration limits :) $\endgroup$ – Soe Karma Jul 31 '12 at 8:57
  • $\begingroup$ Then edit your question, to avoid more answers concerning the antiderivative/indefinite integral... $\endgroup$ – draks ... Jul 31 '12 at 9:04
2
$\begingroup$

$\int\limits_{-\infty}^\infty e^{ikx}dx$ is a (EDIT: scaled by $\frac1{2\pi}$) representation of Dirac's $\delta(k)$ function. For the antiderivative see the other answers...

| cite | improve this answer | |
$\endgroup$
10
$\begingroup$

Integral $$\int_{-a}^{a} e^{ikx}dx=\frac{e^{ikx}}{ik}|_{-a}^{a}=\frac{e^{ika}-e^{-ika}}{ik}=\frac{2i\sin{ka}}{ik}=2a\frac{\sin{ka}}{ka}$$ Now if $a\to\infty$, the term $\frac{\sin{ka}}{ka}=\delta(ka)=\frac{1}{a}\delta(k)$ where $\delta$ is Dirac-delta function.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What if we change the limits as zero to infinity? $\endgroup$ – Zohaib Aarfi Jan 3 '17 at 6:23
  • $\begingroup$ A nice feature for an approximation of the delta function to have is to be in $L^1$. Unfortunately, the sinc function not in $L^1$. $\endgroup$ – robjohn Sep 5 '17 at 18:59
  • $\begingroup$ It should be $\sin(ka)/ka= \pi \delta(ka)$ as $\to \infty$. $\endgroup$ – Cyclone Oct 24 '17 at 17:14
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What if we change the limits as zero to infinity? $\endgroup$ – Zohaib Aarfi Jan 3 '17 at 6:23
  • 2
    $\begingroup$ @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, $\endgroup$ – Raymond Manzoni Jan 3 '17 at 8:52
5
$\begingroup$

$$ \begin{align} \int_{-\infty}^\infty e^{ixy}\overbrace{\ \ \ e^{-\epsilon x^2}\ \ \ }^{\to1}\,\mathrm{d}x &=e^{-\frac{y^2}{4\epsilon}}\color{#C00}{\int_{-\infty}^\infty e^{-\epsilon\left(x-\frac{iy}{2\epsilon}\right)^2}\,\mathrm{d}x}\tag{1}\\ &=\underbrace{\color{#C00}{\sqrt{\frac\pi\epsilon}}\,e^{-\frac{y^2}{4\epsilon}}}_{\to2\pi\delta(y)}\tag{2} \end{align} $$ 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 $0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '17 at 21:05
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 at 6:46
0
$\begingroup$

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:

$$ c=ik $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.