Prove the dirac delta contains all frequencies
some things I've read online show the Fourier transform should look more like a complex exponential.


Don't believe everything you read online!

A signal "contains all frequencies" is a vague description; it can apply to any finite-energy signal $x(t)$ whose Fourier transform $X(f)$ is nonzero for all values of $f$, $-\infty &lt; f &lt; \infty$.  On the other hand, a finite-energy signal necessarily has the property that $|x(t)| \to 0$ as $t \to \infty$ as well as when $t \to -\infty$. Absent this property, $\int_{-\infty}^\infty |x(t)|^2 \, \mathrm dt$ will not converge and so $x(t)$ could not be called a finite-energy signal.

Dirac deltas are not finite-energy signals, but it is common engineering practice to treat them as if they are, and apply all the results in Fourier theory them. That this is legitimate requires more high-power math than I understand and so I, like most other engineers, simply accept the more delicate analyses as being understood. With that, consider that the Fourier transform of $x(t) = \delta(t)$ is $X(f) = 1, -\infty &lt; f &lt; \infty$ and so the Fourier transform is nonzero at all $f, -\infty &lt; f &lt; \infty$, that is,
the Dirac delta "contains all frequencies" as you would like it to. If you would like more generality, then the Fourier transform of $x(t) = \delta(t-\tau)$ is
$$X(f) = \int_{-\infty}^\infty x(t)\exp(-j2\pi ft) \, \mathrm dt = \exp(-j2\pi \tau f)$$
which is a function of $f$ (not $\tau$, that is just a constant), is a complex exponential function of $f$ as you have read online, and is nonzero for all $f, -\infty &lt; f &lt; \infty$, that is, $\delta(t-\tau)$ also "contains all frequencies" just like $\delta(t)$ does.  Exercise: What is the complex exponential in the case $\tau = 0$?


  Either way, would someone be able to clear up for me the way in which the dirac has a broad band frequency spectrum such as white noise?


The definition of white noise is a zero-mean process $\{X(t)\colon -\infty &lt; t &lt; \infty\}$ whose power spectral density is 
$$S_X(f) = \frac{N_0}{2}, -\infty &lt; f &lt; \infty$$
which is not just nonzero but also constant for all $f$. That is, not only does the noise "contain all frequencies" but it contains all frequencies in "equal quantities". If the white noise is filtered through an LTI system with transfer function $H(f)$, then the power spectral density of the output is $\frac{N_0}{2}|H(f)|^2$ and the output noise power is $\frac{N_0}{2}\int_{-\infty}^\infty |H(f)|^2 \,\mathrm df$. In particular, if the filter is an ideal bandpass filter with bandwidth $B$, then the output noise power is $N_0B$ regardless of where the band is in the frequency spectrum.  

In this sense,
the Dirac delta is like white noise: its spectrum $X(f)$ has constant value $1$ for all frequencies (when $\tau = 0$) and more generally, $|X(f)|$ has constant value $1$. The energy spectral density $|X(f)|^2$ of the Dirac delta also has value $1$ for frequencies $f$ (which is similar to the result for the power spectral density of white noise). Also, if the Dirac delta is filtered through an LTI system with transfer function $H(f)$, then the energy of the output signal is $\int_{-\infty}^\infty |H(f)|^2 \,\mathrm df$ which is similar to result for output noise power of the LTI filter when the input is white noise. Finally, if the LTI system is an ideal bandpass filter of bandwidth $B$, then the output is a finite-energy signal of energy $2B$, regardless of what the center frequency of the filter is, which is similar to the comparable result for output noise power from the LTI filter driven by white noise.