photoelectric effect - Number of photons in a range of wavelengths

I need to calculate the number of photons in a beam of light of power $P$. I know that it has constant power $P$ across the range of wavelengths $[\lambda_1,\lambda_2]$. So, for calculating this, I've used a formula that was given in another SE question:

$$N=\frac{1}{h} \int_{\nu_1}^{\nu_2} \frac{1}{\nu} \frac{dE}{d\nu} d\nu$$

It's all fine, and using this I came up with $N=ln(\nu_2/\nu_1)$. But I'm not convinced completely on that formula because I'm not able to derive it from $E=N(\nu)h\nu$.

The answer I get from the formula seems right, but I need proof for that.

Source for the equation: Number of photons

Answer

Power is the amount of energy conveyed per second, so you won't be able to compute the number of photons. Instead, you will compute the number of photons per second. I take $P$ to mean the total beam power within the frequency range from $\nu_1$ to $\nu_2$.

The number of photons per second in a small spectral interval $\delta\nu$ is going to depend on the ratio of beam power in that spectral interval, to the energy per photon in the spectral interval.

The power of the beam is equal to the number of photons per second, divided by the energy per photon. The photons have a range of frequencies, $\nu_1$ through $\nu_2$. The problem states that the power is the same for each frequency within that range.

Let N be the total number of photons per second conveyed by the beam. Let's pick a small frequency range from $\nu_i$ to $\nu_i + \delta\nu$. We can pretend all the photons in that small range have the same frequency, $\nu_i$. So the number of photons per second in that range is $\delta\nu\frac{dP/d\nu}{h\nu_i}$. But $dP/d\nu$ is a constant:

$$dP/d\nu = P/(\nu_2-\nu_1)$$

To find the total number of photons per second in the whole range, we need to add up all the contributions from all the small ranges:

$$N (total photons/sec) = \frac{P}{\nu_2-\nu_1}\sum(\delta\nu\frac{1}{h\nu_i})$$

over all $\nu_i$ in the range. That's just the integral

$$N= \frac{P}{\nu_2-\nu_1}\int_{\nu_1}^{\nu_2} \frac{1}{h\nu} d\nu$$

where $N$ is the number of photons per second within the range from $\nu_1$ to $\nu_2$.

(Hopefully I haven't made any errors in the math. I'm very clumsy with MathJax.)