Ambiguity in the Fourier transform of $f(x)=\cos(ax)$

Question

I am slightly confused about two contradictory answers I am getting with regard to the Fourier transform of the function $f(x)=\cos(ax)$. The first method I used was \begin{align} F(k)&=\int_{-\infty}^{\infty}\cos(ax)e^{-ikx} \ dx\\ &=\frac{1}{2}\int_{-\infty}^{\infty}\left(e^{iax}+e^{-iax}\right)e^{-ikx} \ dx\\ &=\frac{1}{2}\int_{-\infty}^{\infty}e^{-ix(k-a)}+e^{-ix(k+a)} \ dx\\ &=\pi\int_{-\infty}^{\infty}\frac{1}{2\pi}\left(e^{-ix(k-a)}+e^{-ix(k+a)}\right)\\ &=\pi\left(\delta(k-a)+\delta(k+a)\right). \end{align} However, I noticed that a different method was used that resulted in a different answer, namely $$F(k)=\frac{1}{2}(\delta(k-a)+\delta(k+a)).$$ Where am I going wrong in my first method to be out by a factor of $2\pi$.

Answer 1

As discussed in the comments, probably the two answers differ by a factor of $2\pi$ because two different conventions for the definition of Fourier have been used. The second method probably uses
$$ F(k)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(ax)e^{-ikx} \, dx. $$

There are several common conventions for the definition of the Fourier transform, in fact they differ by a multiplicative constant (long story short: some people like $2\pi$ to appear in the definition, some people prefer the definition to be simpler and have the $2\pi$ appear later while studying some properties).