In the context of contour integration:
For positive real values of $\alpha$ the following integral is $$ I(\alpha)=\int_{-\infty}^{\infty}\frac{e^{it\alpha}}{1+t^2}dt=\frac{\pi}{e^{\alpha}} $$ Why does $\alpha$ have to be positive? Does this mean that the following isn't correct? $$ I(-\alpha)=\int_{-\infty}^{\infty}\frac{dt}{(1+t^2)e^{it\alpha}}=\pi e^{\alpha} $$ How and why would the integral be different from the first if this isn't true?
What am I missing here?
On
It's a question of where you close the contour: you can't make the integrals along the other parts of the contour decay if you put it on the wrong side of the real axis.
$$\int_C \frac{e^{i\alpha z}}{1+z^2} \, dz $$ To elaborate, let's use a rectangular contour, to more easily see what's happening. Let it have vertices at $L,L+iK,-L+iK,-L$. Then the integral along this contour decomposes as $$ \int_{-L}^L \frac{e^{i\alpha x}}{1+x^2} \, dx + \int_{0}^{K} \frac{e^{i\alpha(L+iy)} (i \, dy)}{1+(L+iy)^2} + \int_{K}^{-K} \frac{e^{i\alpha(x+iK)}(-dx)}{1+(x+iK)^2} + \int_{K}^{0} \frac{e^{i\alpha(L+iy)} (-i \, dy)}{1+(-L+iy)^2} $$ The first integral is what we want. The second and fourth work in the same way, so I shall just work on the second: notice the absolute value of the integrand is $$ \left\lvert \frac{e^{i\alpha L-\alpha y} i}{1+(L+iy)^2} \right\rvert = \frac{e^{-\alpha y}}{((1+L^2-y^2)^2+4y^2L^2)^{1/2}}. $$ The denominator is bounded below by a multiple of $1/L$, and so the integral is bounded above by $$ \frac{c}{L} \int_0^{\infty} e^{-\alpha y} \, dy = \frac{c}{\alpha L} \to 0 $$ as $L \to \infty$, providing that $\alpha>0$: if it was not, the integrand would not decrease for large $y$, even with a polynomial on the bottom.
That's leaves one integral. The third has numerator bounded above by $e^{-\alpha K}$ in the same way, and since $$ \frac{1}{((1+x^2-K^2)^2+4x^2K^2)^{1/2}} $$ is has finite integral over $(-\infty,\infty)$ if $K$ is large enough (note the the denominator has no roots and goes approximately like $x^{-2}$ as $x \to \infty$), so this integral decays at least as fast as $e^{-\alpha K}$ as $K \to \infty$. Again, this would not work if $\alpha$ were not larger than zero.
Then you do the usual thing with evaluating the residue at $i$, and find the value of the integral is $\pi e^{-\alpha}$.
Now, that doesn't work for $\alpha<0$: the absolute value of the integrand grows as $\Im(z) \to \infty$ in that case. So we have to use a different contour: take one in the lower half-plane, vertices at $L,L-iK,-L-iK,-L$. For the integrand does decay as $\Im(z) \to -\infty$, and we can use the same argument, and then use the residue of the pole inside the contour to compute the integral: in particular, the pole in question is now $-i$, where the residue is $$ \frac{e^{(-i)i\alpha}}{-2i} = -\frac{e^{\alpha}}{2i}, $$ so the integral is $-\pi e^{\alpha}$. Well, almost: the contour I gave is traversed in the opposite direction to normal (i.e. clockwise rather than anticlockwise), so you pick up an extra minus sign and the answer is $\pi e^{\alpha}$.
Take conjugate on both sides of $\int_{-\infty}^{\infty}\frac{e^{it\alpha}}{1+t^2}dt=\frac{\pi}{e^{\alpha}}$, you get exactly $\int_{-\infty}^{\infty}\frac{e^{-it\alpha}}{1+t^2}dt=\frac{\pi}{e^{\alpha}},\space$ therefore $I(-\alpha)=I(\alpha)$.
Another way to think of it is, changing variable $u=-t$, you have:
$\int_{-\infty}^{\infty}\frac{e^{it\alpha}}{1+t^2}dt\overset{u=-t}{=}\int_{+\infty}^{-\infty}\frac{e^{-iu\alpha}}{1+(-u)^2}d(-u)=\int_{-\infty}^{\infty}\frac{e^{-iu\alpha}}{1+(u)^2}d(u)$.