Show that $\int_{\mathbb R} \frac{\alpha ^2}{\alpha ^2 +\lambda ^2}=\frac{\pi}{2\alpha }$

Question

Fix that $\alpha >0$ and consider the function $f(x)=e^{-2\pi \alpha |x|} $,$x\in \mathbb{R}$
1.Compute the fourier transform of $f$.
2.Show that $\int_{\mathbb R} \frac{\alpha ^2}{\alpha ^2 +w ^2}=\frac{\pi}{2\alpha }$

We have that $\hat{f}(w)=\int_{\mathbb R}f(x)e^{-2\pi iwx}dx$ and i find that $\hat{f}(w)= \frac {\alpha}{\pi(\alpha ^2 + w^2)}.$
For the second question we have that $$\hat{f}(w)= \frac {\alpha}{\pi(\alpha ^2 + w^2)}$$ then $$\hat{\hat{f}}(w)= F(\frac {\alpha}{\pi(\alpha ^2 + w^2)})$$ by the inverse theorem we have $$\hat{\hat{f}}(w)=f(-x)$$ and $$F(\frac {\alpha}{\pi(\alpha ^2 + w^2)})=\alpha \pi F(\frac{\alpha ^2}{\alpha ^2 + w^2}) $$ we finally we got $$e^{-2\pi \alpha |x|}=\alpha \pi \int_{\mathbb R} \frac{\alpha ^2}{\alpha ^2 +w ^2}e^{-2\pi iwx}dx$$ we take $x=0$ we got $$ \int_{\mathbb R} \frac{\alpha ^2}{\alpha ^2 +w ^2}= \frac{1}{\alpha \pi }$$ I got wrong result.
Any feedback is appreciated!

Answer 1

Another quick way, due to the definition of Fourier transform, is $$ \int_\mathbb{R}\frac{\alpha^2}{\alpha^2+w^2}\mathrm{d}w = \left.\int_\mathbb{R}\frac{\alpha^2}{\alpha^2+w^2}e^{iwx}\mathrm{d}w\right|_{x=0} = \pi\alpha\,\mathscr{F}^{-1}\left[\frac{\alpha}{\pi(\alpha^2+w^2)}\right](0) = \left.\pi\alpha e^{-2\pi\alpha|x|}\right|_{x=0} = \pi\alpha $$

Answer 2

A quick solution to $(2)$ is the following. By the change of variable $w = \alpha x$, $$ \int_{\Bbb R} \frac{\alpha^2}{\alpha^2+w^2} \,\mathrm d w = \int_{\Bbb R} \frac{\alpha^2}{\alpha^2+\alpha^2x^2} \,\alpha\,\mathrm d x = \alpha \int_{\Bbb R} \frac{1}{1+x^2}\,\mathrm d x \\ = \alpha \,\Big[\arctan\Big]_{-\infty}^\infty = \alpha\,\pi. $$