# Derivation of $\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)\cos(t\xi)dt$

Problem: Consider the Fourier transform: $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{-it\xi}dt\quad\text{and}\quad f\in{L_1}_{(\mathbb{R})}$$

Now it was defined to me on ${L_1}_{(\mathbb{R})}$ $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)\cos(t\xi)dt$$

Given the Euler identity $e^{it\xi}=\cos(t\xi)+i\sin(t\xi)$, I do not know how the $i\sin(t\xi)$ vanished.

Question:

How did it happen?

Edit: Urgent

I only noticed it now, the transform defined is the following:

$$\hat{f}(\xi)=\frac{\sqrt{2}}{\sqrt{\pi}}\int_\limits{0}^{\infty}f(t)\cos(t\xi)dt$$

Apologies!

• f(t) even function? or the transform is real. – herb steinberg Apr 28 '18 at 16:52
• @herbsteinberg Please check the update! – Pedro Gomes Apr 28 '18 at 16:58
• Yeah, the only way this would work would be that your function is even. – Joel Apr 28 '18 at 17:10
• Even so how would the $i\sin(t\xi)$ vanish? – Pedro Gomes Apr 28 '18 at 17:18
• If $f$ is even then $f(t)\sin(\xi t)$ is odd, so the integral over symmetric intervals is zero. – Joel Apr 28 '18 at 17:28

That identity can only work for even functions. Recall that if a function $a(t)$ is odd, then $\int_{-t}^t a(\tau) d\tau = 0$.

Thus, for even $f$, $$\int_{-\infty}^\infty f(t) \sin(\xi t) dt = 0$$ since $f(t) \sin(\xi t)$ is odd for all $\xi$. (Product of an even function with an odd function is odd.)

Now, suppose that $\hat f(\xi) = \int_{-\infty}^\infty f(t) \cos(\xi t) dt.$ We have that $\hat f$ is even, since $\cos$ is even.

If we want to recover $f$, then we can apply the inverse Fourier transform: $$f(t) = \int_{-\infty}^\infty \hat f(\xi) e^{i\xi t} d\xi = \int_{-\infty}^\infty \hat f(\xi) \cos(\xi t) d\xi + i \int_{-\infty}^\infty \hat f(\xi) \sin(\xi t) d\xi.$$ Just as before $\hat f$ even means that $\int_{-\infty}^\infty \hat f(\xi) \sin(\xi t) d\xi = 0$. Hence, $f(t) = \int_{-\infty}^\infty \hat f(\xi) \cos(\xi t) d\xi$.

Finally, we have that $f$ is even, since $\cos$ is even.

Thus, $\hat f(\xi) = \int_{-\infty}^\infty f(t) \cos(\xi t) dt$ iff $f$ is even.

As for your edit, I did neglect the constants above. This is often done for expediency, and I can see how it can be confusing.

Taking $\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{-\xi t} dt$, we already established that $\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) \cos(\xi t) dt$.

Now for even functions, say $b(t)$, we have $\int_{-a}^a b(t) dt = 2 \int_0^a b(t) dt$. Thus, $$\hat f(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) \cos(\xi t) dt = \frac{2}{\sqrt{2\pi}} \int_{0}^\infty f(t) \cos(\xi t) dt = \frac{\sqrt2}{\sqrt{\pi}} \int_{0}^\infty f(t) \cos(\xi t) dt$$

• We had to use $f$ in $L^1$ to recover $f$ from its Fourier transform. – Joel Apr 28 '18 at 18:01
• Thanks for the update! I have already reviewed the topics. I know $\sin(x)$ to be a odd function so $\int_\limits{-N}^{N}\sin(x)=0$. Since the integral interval is now $[0,\infty)$. Why would the odd function vanish? – Pedro Gomes Apr 28 '18 at 20:43
• Apparently the function is assumed to be even. In that case the integral $(-\infty,\infty)=$ twice the integral $(0,\infty)$. – herb steinberg Apr 29 '18 at 15:03