Let $f(x) = \frac{1}{1+|x|}$. I just showed that $f \in L^2(\Bbb{R})$, and I'm being asked to find the $L^2$-norm of the Fourier transform. However, the formula I'm given for the Fourier transform only works for $L^1$ functions. More precisely, the lecture notes I am using defines
$$g(\lambda) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(t) e^{-i\lambda t} dt$$
to be the Fourier transform of some function $f$, and it is remarked that this integral converges for $f \in L^1(\Bbb{R})$, but $f$ as defined above is not an $L^1(\Bbb{R})$-function.
Ignoring this for a moment, I tried to blindly apply the formula and I got the following:
$$\int_{-\infty}^{\infty} f(t) e^{-i\lambda t} dt = \int_{-\infty}^{0} \frac{1}{1-t} e^{-i \lambda t} dt + \int_{0}^{\infty} \frac{1}{1+t} e^{- i \lambda t} dt$$
$$= \int_{0}^{\infty} \frac{1}{1+t} e^{i \lambda t} dt + \int_{0}^{\infty} \frac{1}{1+t} e^{- i \lambda t} dt$$
$$2 \int_{0}^{\infty} \frac{1}{1+t} Re(e^{i \lambda t})$$
$$=2 \int_{0}^{\infty} \frac{\cos (\lambda t)}{1+t} dt$$ But I am not really sure what to do with this integral. In fact, doesn't this diverge for $\lambda = 0$, for example? What am I doing wrong?