## Fourier Inversion Theorem

## Introduction

The Fourier Transform and it's inverse are defined (in my convention) as this:

\[ F(\omega) = \int_{-\infty}^{\infty} f(t) \exp(i \omega t) dt\]

\[ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega \]

What I set out to show is where that simple $2\pi$ factor comes from - which turns out to not be so simple. It turns out that as part of this proof, I will also need to show the fairly often encountered integral:

\[ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx} dx = \delta(k) \]

Where $\delta(x)$ is the Dirac delta function. I assume knowledge at least of how the delta function works (how it's defined and it's sifting property are the most important).

Also, note that I will not be very mathematically rigorous in my treatment of this problem. To do so would require talking about the theory of distributions (for the Dirac delta function), as well as measure theory, and convergence of the Fourier Transform for certain functions which are themselves not absolutely convergent over the whole domain (like $\sin(\omega_0 t)$). I pretty much know nothing of these topics, so I will invoke these tools when necessary with no justification. Thankfully the details of these proofs aren't intimately involved in getting the $2\pi$ result.

\[ F(\omega) = \int_{-\infty}^{\infty} f(t) \exp(i \omega t) dt\]

\[ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega \]

What I set out to show is where that simple $2\pi$ factor comes from - which turns out to not be so simple. It turns out that as part of this proof, I will also need to show the fairly often encountered integral:

\[ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx} dx = \delta(k) \]

Where $\delta(x)$ is the Dirac delta function. I assume knowledge at least of how the delta function works (how it's defined and it's sifting property are the most important).

Also, note that I will not be very mathematically rigorous in my treatment of this problem. To do so would require talking about the theory of distributions (for the Dirac delta function), as well as measure theory, and convergence of the Fourier Transform for certain functions which are themselves not absolutely convergent over the whole domain (like $\sin(\omega_0 t)$). I pretty much know nothing of these topics, so I will invoke these tools when necessary with no justification. Thankfully the details of these proofs aren't intimately involved in getting the $2\pi$ result.

## Fourier Transform And Its Inverse

I take the general form of the Fourier Transform and it's inverse to be assumed, but I do not assume anything about the constants out in front.\[ F(\omega) = A \int_{-\infty}^{\infty} f(t) \exp(i\omega t) dt \]

\[ f(t) = B \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega\]

In order for the inverse transform to appropriately undo the forward transform, you'd expect a serial application of the forward and then inverse transform to yield you the same function you started with. Otherwise, calling this a transform wouldn't make a whole bunch of sense.

\[ f(t) = B \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega = B \int_{-\infty}^{\infty}\left[ A\int_{-\infty}^{\infty} f(\tau) \exp(i\omega\tau)d\tau\right] \exp(-i\omega t) d\omega\]

If I interchange the order of integration (which I assume to be a legal maneuver)

\[ f(t) = \int_{-\infty}^{\infty} \left[ AB \int_{-\infty}^{\infty} \exp(i \omega (t-\tau)) d\omega \right] f(\tau) d\tau \]

The only way I can force $f(t)$ on the left hand side to equal the right hand side for all time $t$ is to use the Dirac delta function. If the quantity in brackets was $\delta(t-\tau)$, then the sifting property of the delta function could kill the integral, and we'd recover the required $f(t)$ on the right hand side for all $t$. Therefore, the problem has become proving this:

\[ AB \int_{-\infty}^{\infty} \exp(i\omega(t-\tau))d\omega = \delta(t-\tau) \]

To make sense of this, we must discuss more properties of the delta function.

\[ f(t) = B \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega\]

In order for the inverse transform to appropriately undo the forward transform, you'd expect a serial application of the forward and then inverse transform to yield you the same function you started with. Otherwise, calling this a transform wouldn't make a whole bunch of sense.

\[ f(t) = B \int_{-\infty}^{\infty} F(\omega) \exp(-i\omega t) d\omega = B \int_{-\infty}^{\infty}\left[ A\int_{-\infty}^{\infty} f(\tau) \exp(i\omega\tau)d\tau\right] \exp(-i\omega t) d\omega\]

If I interchange the order of integration (which I assume to be a legal maneuver)

\[ f(t) = \int_{-\infty}^{\infty} \left[ AB \int_{-\infty}^{\infty} \exp(i \omega (t-\tau)) d\omega \right] f(\tau) d\tau \]

The only way I can force $f(t)$ on the left hand side to equal the right hand side for all time $t$ is to use the Dirac delta function. If the quantity in brackets was $\delta(t-\tau)$, then the sifting property of the delta function could kill the integral, and we'd recover the required $f(t)$ on the right hand side for all $t$. Therefore, the problem has become proving this:

\[ AB \int_{-\infty}^{\infty} \exp(i\omega(t-\tau))d\omega = \delta(t-\tau) \]

To make sense of this, we must discuss more properties of the delta function.

## Dirac Delta Properties

The first thing we need to understand about the delta function is that it isn't a "function" in the usual sense of the word. Formally, it would be a distribution, but I don't want to delve into this. Instead, just think of it as a "function" that only makes sense as part of an integral. We note that as long as the bounds of an integral pass over the spike in the delta function (which occurs when it's argument is zero), then it equals 1, otherwise zero. To illustrate this:

\[ \int_a^b \delta(x) dx = \begin{cases} 1 & a<0<b \\ 0 & a,b<0 \\ 0 & a,b>0 \\ -1 & b<0<a \end{cases} \]

Note that I have avoided the $a=0$ or $b=0$ case. In these cases, since bound of the integral diverges, it's tough to assign a number. But then again, since the delta function is clearly an even "function", one might expect that an integral over just "half" of the spike would result in a factor of $\frac{1}{2}$. I can't really

With all of these cases, we could look at the $\text{sign}(x)$ function and try to recast our piecewise function in terms of the $\text{sign}(x)$ function.

\[ \int_a^b \delta(x) dx =\frac{\text{sign}(b) - \text{sign}(a)}{2} \]

Which, if you work through the different cases, you'll see these are equivalent definitions. Therefore, if we take the result from the last section and integrate it also with these arbitrary bounds, we'd get:

\[ AB \int_a^b \left( \int_{-\infty}^{\infty} d\omega \exp(i\omega(t-\tau)) \right) d(t-\tau) = \frac{\text{sign}(b)-\text{sign}(a)}{2}\]

\[ \int_a^b \delta(x) dx = \begin{cases} 1 & a<0<b \\ 0 & a,b<0 \\ 0 & a,b>0 \\ -1 & b<0<a \end{cases} \]

Note that I have avoided the $a=0$ or $b=0$ case. In these cases, since bound of the integral diverges, it's tough to assign a number. But then again, since the delta function is clearly an even "function", one might expect that an integral over just "half" of the spike would result in a factor of $\frac{1}{2}$. I can't really

*prove*this necessarily, but hopefully this makes intuitive sense.With all of these cases, we could look at the $\text{sign}(x)$ function and try to recast our piecewise function in terms of the $\text{sign}(x)$ function.

\[ \int_a^b \delta(x) dx =\frac{\text{sign}(b) - \text{sign}(a)}{2} \]

Which, if you work through the different cases, you'll see these are equivalent definitions. Therefore, if we take the result from the last section and integrate it also with these arbitrary bounds, we'd get:

\[ AB \int_a^b \left( \int_{-\infty}^{\infty} d\omega \exp(i\omega(t-\tau)) \right) d(t-\tau) = \frac{\text{sign}(b)-\text{sign}(a)}{2}\]

## The First Integral - Sign Function

Taking the previous expression, changing the $t-\tau$ variable to $x$, and $\omega$ to $k$ (as a matter of taste), we get:

\[ AB \int_a^b dx \int_{-\infty}^{\infty}dk \exp(ikx) = \frac{\text{sign}(b)-\text{sign}(a)}{2}\]

Let's play with the interior integral first, which we do by breaking it into 2 pieces:

\[ \int_{-\infty}^{\infty} e^{ikx} dk = \int_0^{\infty} e^{ikx} dk + \int_{-\infty}^0 e^{ikx} dk\]

Changing variables $k' = -k$ for just the second integral, we find:

\[ = \int_0^{\infty} e^{ikx}dk + \int_{\infty}^0 e^{-ik'x} (-dk') = \int_0^{\infty} \left( e^{ikx} + e^{-ikx} \right) dk\]

Noting of course that the quantity in parentheses is just $2\cos(kx)$, we can rewrite our integral (with another interchange of integration order) to get:

\[ 4AB \int_0^{\infty} dk \int_a^b \cos(kx) dx = \text{sign}(b)-\text{sign}(a)\]

Next, we work on the (new) interior integral, which is simply the integral over a finite domain of $\cos(kx)$, which is quite simple. The result:

\[ 4AB \int_0^{\infty} \left[ \frac{sin(kx)}{k}\right]_a^b dk = \text{sign}(b)-\text{sign}(a) = 4AB \int_0^{\infty} \frac{\sin(kb)-\sin(ka)}{k} dk\]

The next trick is to change the bounds on the $k$ integral to finite numbers, and take the limit. This gives us:

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \int_{\epsilon}^{\lambda} \frac{\sin(kb)-\sin(ka)}{k} dk\]

Let's break this into 2 integrals.

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \left[ \int_{\epsilon}^{\lambda} \frac{\sin(kb)}{k}dk - \int_{\epsilon}^{\lambda} \frac{\sin(ka)}{k} dk\right]\]

A change of variables on each of these integrals such that $u=kb$ and $v=ka$, gives:

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \left[ \int_{\epsilon/b}^{\lambda/b} \frac{\sin(u)}{u}du - \int_{\epsilon/a}^{\lambda/a} \frac{\sin(v)}{v} dv\right]\]

Now, note the fact that the integrands of both functions are the same - all that changes are the bounds on the integral. Suppose briefly that $a,b>0$. In this case, the limits take both integrals from $0^+$ to $+\infty$, and when we subtract, we arrive at 0. Separately, consider $a,b<0$. In this case, both integrals now range from $0^-$ to $-\infty$, which will also disappear when they subtract. However, what if $a$ and $b$ have different signs? Let's suppose $b>0$ and $a<0$. Taking these limits we'd get:

\[ \int_{0^+}^{+\infty} \frac{\sin u}{u} du - \int_{0^-}^{-\infty} \frac{\sin v}{v} dv \]

Yet another change of variables $v=-u$ allows us to change the second integral into the first, but the sign out in front changes, leaving us with:

\[ 2 \int_{0^+}^{+\infty} \frac{\sin u}{u} du \]

Therefore, we can see that, summarizing these different cases, we have the same integral of $\frac{\sin u}{u}$, but the coefficient out in front will either be $0$ or $2$, corresponding perfectly with $\text{sign}(b)-\text{sign}(a)$. Plugging this all back into our earlier equation, we arrive at:

\[ 4AB (\text{sign}(b)-\text{sign}(a) ) \int_0^{\infty} \frac{\sin u}{u} du = \text{sign}(b) - \text{sign}(a) \]

Cancelling out the two $\text{sign}$ functions on either side gives us:

\[ \int_0^{\infty} \frac{\sin u}{u} du = \frac{1}{4AB}\]

Therefore, if we could evaluate the integral on the left, we could solve for the product $AB$, which, as a reminder, are the constants that go out in front of the Fourier Transforms - which is what we set out to find in the first place.

\[ AB \int_a^b dx \int_{-\infty}^{\infty}dk \exp(ikx) = \frac{\text{sign}(b)-\text{sign}(a)}{2}\]

Let's play with the interior integral first, which we do by breaking it into 2 pieces:

\[ \int_{-\infty}^{\infty} e^{ikx} dk = \int_0^{\infty} e^{ikx} dk + \int_{-\infty}^0 e^{ikx} dk\]

Changing variables $k' = -k$ for just the second integral, we find:

\[ = \int_0^{\infty} e^{ikx}dk + \int_{\infty}^0 e^{-ik'x} (-dk') = \int_0^{\infty} \left( e^{ikx} + e^{-ikx} \right) dk\]

Noting of course that the quantity in parentheses is just $2\cos(kx)$, we can rewrite our integral (with another interchange of integration order) to get:

\[ 4AB \int_0^{\infty} dk \int_a^b \cos(kx) dx = \text{sign}(b)-\text{sign}(a)\]

Next, we work on the (new) interior integral, which is simply the integral over a finite domain of $\cos(kx)$, which is quite simple. The result:

\[ 4AB \int_0^{\infty} \left[ \frac{sin(kx)}{k}\right]_a^b dk = \text{sign}(b)-\text{sign}(a) = 4AB \int_0^{\infty} \frac{\sin(kb)-\sin(ka)}{k} dk\]

The next trick is to change the bounds on the $k$ integral to finite numbers, and take the limit. This gives us:

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \int_{\epsilon}^{\lambda} \frac{\sin(kb)-\sin(ka)}{k} dk\]

Let's break this into 2 integrals.

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \left[ \int_{\epsilon}^{\lambda} \frac{\sin(kb)}{k}dk - \int_{\epsilon}^{\lambda} \frac{\sin(ka)}{k} dk\right]\]

A change of variables on each of these integrals such that $u=kb$ and $v=ka$, gives:

\[ \lim_{\lambda\rightarrow\infty} \lim_{\epsilon\rightarrow 0} 4AB \left[ \int_{\epsilon/b}^{\lambda/b} \frac{\sin(u)}{u}du - \int_{\epsilon/a}^{\lambda/a} \frac{\sin(v)}{v} dv\right]\]

Now, note the fact that the integrands of both functions are the same - all that changes are the bounds on the integral. Suppose briefly that $a,b>0$. In this case, the limits take both integrals from $0^+$ to $+\infty$, and when we subtract, we arrive at 0. Separately, consider $a,b<0$. In this case, both integrals now range from $0^-$ to $-\infty$, which will also disappear when they subtract. However, what if $a$ and $b$ have different signs? Let's suppose $b>0$ and $a<0$. Taking these limits we'd get:

\[ \int_{0^+}^{+\infty} \frac{\sin u}{u} du - \int_{0^-}^{-\infty} \frac{\sin v}{v} dv \]

Yet another change of variables $v=-u$ allows us to change the second integral into the first, but the sign out in front changes, leaving us with:

\[ 2 \int_{0^+}^{+\infty} \frac{\sin u}{u} du \]

Therefore, we can see that, summarizing these different cases, we have the same integral of $\frac{\sin u}{u}$, but the coefficient out in front will either be $0$ or $2$, corresponding perfectly with $\text{sign}(b)-\text{sign}(a)$. Plugging this all back into our earlier equation, we arrive at:

\[ 4AB (\text{sign}(b)-\text{sign}(a) ) \int_0^{\infty} \frac{\sin u}{u} du = \text{sign}(b) - \text{sign}(a) \]

Cancelling out the two $\text{sign}$ functions on either side gives us:

\[ \int_0^{\infty} \frac{\sin u}{u} du = \frac{1}{4AB}\]

Therefore, if we could evaluate the integral on the left, we could solve for the product $AB$, which, as a reminder, are the constants that go out in front of the Fourier Transforms - which is what we set out to find in the first place.

## The Second Integral - Sine Function

There is a function called the $\text{Si}(x)$, or "sine integral" function, whose definition is:

\[ \text{Si}(x) = \int_0^{x} \frac{\sin u}{u} du \]

So if we were already familiar with this function, we would just find it's limit as $x\rightarrow\infty$ and be done. But, at least personally, I've never needed this function, so appealing to it's asymptotic limit feels like cheating to me. That said, let's try another trick.

Suppose instead we have this function:

\[ I(x) = \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du\]

Where $x$ is a positive, real number (negative would cause the integral to blow up, and complex adds unnecessary complication). Without getting bogged down in the mathematical rigor, let's appeal to the well-behaved-ness of the functions to take a derivative on both sides with respect to $x$. Interchanging the integral with the derivative, we get:

\[ \frac{dI(x)}{dx} = \int_0^{\infty} \frac{\partial}{\partial x} \left(\frac{\sin u}{u} \exp(-ux) \right) du = -\int_0^{\infty} \sin u \exp(-ux) du\]

This integral now is worked out more easily. We could integrate by parts cleverly and get the result, but I find it's easier to rewrite $\sin u = \frac{\exp(iu)-\exp(-iu)}{2i}$, and simply integrate two exponentials.

\[ \frac{dI(x)}{dx} = \frac{-1}{2i} \int_0^{\infty} e^{u(i-x)} - e^{u(-i-x)} du = \frac{-1}{2i} \left[ \frac{e^{u(i-x)}}{i-x} - \frac{e^{u(-i-x)}}{-i-x}\right]_0^{\infty}\]

Thanks to the $x>0$ condition, the $u\rightarrow\infty$ boundary disappears because of the exponential decay applied by $\exp(-ux)$. The lower bound is all that remains, giving us:

\[ \frac{d I(x)}{dx} = \frac{+1}{2i} \left( \frac{1}{i-x} - \frac{1}{-i-x}\right) = \frac{1}{2i} \frac{i-x+i+x}{(i-x)(i+x)} = \frac{-1}{1+x^2}\]

Therefore, we've solved this integral! But to do so, we needed to differentiate, meaning we still need to evaluate one more integral in order to solve the problem. The indefinite integral is:

\[ I(x) = \int \frac{dI(x)}{dx} dx = -\int \frac{1}{1+x^2} dx\]

Some might recognize this as $\arctan(x)$, but I'd like to prove this briefly. Let's introduce a triangle:

\[ \text{Si}(x) = \int_0^{x} \frac{\sin u}{u} du \]

So if we were already familiar with this function, we would just find it's limit as $x\rightarrow\infty$ and be done. But, at least personally, I've never needed this function, so appealing to it's asymptotic limit feels like cheating to me. That said, let's try another trick.

Suppose instead we have this function:

\[ I(x) = \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du\]

Where $x$ is a positive, real number (negative would cause the integral to blow up, and complex adds unnecessary complication). Without getting bogged down in the mathematical rigor, let's appeal to the well-behaved-ness of the functions to take a derivative on both sides with respect to $x$. Interchanging the integral with the derivative, we get:

\[ \frac{dI(x)}{dx} = \int_0^{\infty} \frac{\partial}{\partial x} \left(\frac{\sin u}{u} \exp(-ux) \right) du = -\int_0^{\infty} \sin u \exp(-ux) du\]

This integral now is worked out more easily. We could integrate by parts cleverly and get the result, but I find it's easier to rewrite $\sin u = \frac{\exp(iu)-\exp(-iu)}{2i}$, and simply integrate two exponentials.

\[ \frac{dI(x)}{dx} = \frac{-1}{2i} \int_0^{\infty} e^{u(i-x)} - e^{u(-i-x)} du = \frac{-1}{2i} \left[ \frac{e^{u(i-x)}}{i-x} - \frac{e^{u(-i-x)}}{-i-x}\right]_0^{\infty}\]

Thanks to the $x>0$ condition, the $u\rightarrow\infty$ boundary disappears because of the exponential decay applied by $\exp(-ux)$. The lower bound is all that remains, giving us:

\[ \frac{d I(x)}{dx} = \frac{+1}{2i} \left( \frac{1}{i-x} - \frac{1}{-i-x}\right) = \frac{1}{2i} \frac{i-x+i+x}{(i-x)(i+x)} = \frac{-1}{1+x^2}\]

Therefore, we've solved this integral! But to do so, we needed to differentiate, meaning we still need to evaluate one more integral in order to solve the problem. The indefinite integral is:

\[ I(x) = \int \frac{dI(x)}{dx} dx = -\int \frac{1}{1+x^2} dx\]

Some might recognize this as $\arctan(x)$, but I'd like to prove this briefly. Let's introduce a triangle:

Simple trig tells us $\tan \theta = x$, or, $\theta = \arctan(x)$. Let's implicitly differentiate the first, giving us:
\[ \frac{d\theta}{dx} \frac{d\tan\theta}{d\theta} = 1\] Since I don't bother to remember the derivative of tangent, let's just derive it real quick via the quotient rule, and solve to get $\frac{d\theta}{dx}$. |

\[ \frac{d}{d\theta} \frac{\sin\theta}{\cos\theta} = \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} = \sec^2\theta ~~~\rightarrow~~~\frac{d\theta}{dx} = \cos^2\theta\]

Using the triangle, we can express this RHS in terms of $x$. Looking at the adjacent and hypotenuse sides, we get:

\[ \frac{d\theta}{dx} = \left( \frac{1}{\sqrt{1+x^2}}\right)^2 = \frac{1}{1+x^2}\]

Therefore, we have found:

\[ \frac{d}{dx} \arctan x = \frac{1}{1+x^2}\]

Therefore, integrating both sides is exactly what we need to do to get $I(x)$. The integral of a derivative undo one another (fundamental theorem of calculus), and, with our undetermined constant, we get:

\[ I(x) = - \arctan(x) + C\]

So, if we knew what the value of $I(x)$ was for a particular value of $x$, we could use this to get $I(x)$ for any $x$. Returning to our definition of $I(x)$:

\[ I(x) = \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du\]

If we let $x\rightarrow\infty$, then the exponential decay destroys the integrand, leaving $I(x)$ to be $0$. In other words:

\[ I(x\rightarrow\infty) = - \arctan(\infty) + C = 0 ~~~\rightarrow~~~ C = \arctan(\infty) = \frac{\pi}{2}\]

\[ \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du = \frac{\pi}{2} - \arctan(x)\]

Now, for our problem, we need to let $x\rightarrow 0$, so that we may finally arrive at the sine integral we needed in the first place. Since we know $\arctan(0)=0$, we get:

\[ \int_0^{\infty} \frac{\sin u}{u} du = \frac{\pi}{2}\]

Now we can combine with the results from the last section to arrive at the final answer:

\[ AB = \frac{1}{4}\left(\frac{2}{\pi}\right) = \frac{1}{2\pi} \]

Which is the result we wanted!

Using the triangle, we can express this RHS in terms of $x$. Looking at the adjacent and hypotenuse sides, we get:

\[ \frac{d\theta}{dx} = \left( \frac{1}{\sqrt{1+x^2}}\right)^2 = \frac{1}{1+x^2}\]

Therefore, we have found:

\[ \frac{d}{dx} \arctan x = \frac{1}{1+x^2}\]

Therefore, integrating both sides is exactly what we need to do to get $I(x)$. The integral of a derivative undo one another (fundamental theorem of calculus), and, with our undetermined constant, we get:

\[ I(x) = - \arctan(x) + C\]

So, if we knew what the value of $I(x)$ was for a particular value of $x$, we could use this to get $I(x)$ for any $x$. Returning to our definition of $I(x)$:

\[ I(x) = \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du\]

If we let $x\rightarrow\infty$, then the exponential decay destroys the integrand, leaving $I(x)$ to be $0$. In other words:

\[ I(x\rightarrow\infty) = - \arctan(\infty) + C = 0 ~~~\rightarrow~~~ C = \arctan(\infty) = \frac{\pi}{2}\]

\[ \int_0^{\infty} \frac{\sin u}{u} \exp(-ux) du = \frac{\pi}{2} - \arctan(x)\]

Now, for our problem, we need to let $x\rightarrow 0$, so that we may finally arrive at the sine integral we needed in the first place. Since we know $\arctan(0)=0$, we get:

\[ \int_0^{\infty} \frac{\sin u}{u} du = \frac{\pi}{2}\]

Now we can combine with the results from the last section to arrive at the final answer:

\[ AB = \frac{1}{4}\left(\frac{2}{\pi}\right) = \frac{1}{2\pi} \]

Which is the result we wanted!

## The Final Answer

After all this work, we have derived two main results. The first:

\[ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-x')} dk = \delta(x-x') \]

Which is an often quoted result in physics textbooks. The second thing, we have determined that after applying the forward and backward Fourier Transform, we must pick up a factor of $\frac{1}{2\pi}$ somehow, though the arrangement is arbitrary. Two common choices are $A=B=\frac{1}{\sqrt{2\pi}}$, which looks nice since the forward and the backward transform therefore differ only by a complex conjugate, and $A=1, B=\frac{1}{2\pi}$, which is the form I'm most used to seeing in textbooks. Ultimately the choice of $A$ and $B$ is arbitrary, so long as their product is kept constant at $\frac{1}{2\pi}$.

Who would have thought a simple factor of $2\pi$ would be so difficult to uncover?

Also, I'd like to thank a few sites that helped with these derivations. This source helped with the first half of the derivation, this forum thread gave (numerous) ways to handle the Sine Integral derivation, and this page introduces some very cool integration techniques, one of which was employed in handling the Sine Integral. Also, shout out to my high school calculus teacher for the trig / inverse trig derivative trick - thanks Mr. King!

08/14/2014

\[ \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ik(x-x')} dk = \delta(x-x') \]

Which is an often quoted result in physics textbooks. The second thing, we have determined that after applying the forward and backward Fourier Transform, we must pick up a factor of $\frac{1}{2\pi}$ somehow, though the arrangement is arbitrary. Two common choices are $A=B=\frac{1}{\sqrt{2\pi}}$, which looks nice since the forward and the backward transform therefore differ only by a complex conjugate, and $A=1, B=\frac{1}{2\pi}$, which is the form I'm most used to seeing in textbooks. Ultimately the choice of $A$ and $B$ is arbitrary, so long as their product is kept constant at $\frac{1}{2\pi}$.

Who would have thought a simple factor of $2\pi$ would be so difficult to uncover?

Also, I'd like to thank a few sites that helped with these derivations. This source helped with the first half of the derivation, this forum thread gave (numerous) ways to handle the Sine Integral derivation, and this page introduces some very cool integration techniques, one of which was employed in handling the Sine Integral. Also, shout out to my high school calculus teacher for the trig / inverse trig derivative trick - thanks Mr. King!

08/14/2014