We want to understand the integral

$\displaystyle \int_{-\infty}^\infty \frac{\mathrm{d}t}{(1 + t^2)^n}. \ \ \ \ \ (1)$

Although fattybake mentions the residue theorem, we won't use that at all. Instead, we will be very clever.

We will do a technique that was once very common (up until the 1910s or so), but is much less common now: let's multiply by ${\displaystyle \Gamma(n) = \int_0^\infty u^n e^{-u} \frac{\mathrm{d}u}{u}}$. This yields

$\displaystyle \int_0^\infty \int_{-\infty}^\infty \left(\frac{u}{1 + t^2}\right)^n e^{-u}\mathrm{d}t \frac{\mathrm{d}u}{u} = \int_{-\infty}^\infty \int_0^\infty \left(\frac{u}{1 + t^2}\right)^n e^{-u} \frac{\mathrm{d}u}{u}\mathrm{d}t, \ \ \ \ \ (2)$

where I interchanged the order of integration because everything converges really really nicely. Do a change of variables, sending ${u \mapsto u(1+t^2)}$. Notice that my nicely behaving measure ${\mathrm{d}u/u}$ completely ignores this change of variables, which is why I write my ${\Gamma}$ function that way. Also be pleased that we are squaring ${t}$, so that this is positive and doesn't mess with where we are integrating. This leads us to

$\displaystyle \int_{-\infty}^\infty \int_0^\infty u^n e^{-u + -ut^2} \frac{\mathrm{d}u}{u}\mathrm{d}t = \int_0^\infty \int_{-\infty}^\infty u^n e^{-u + -ut^2} \mathrm{d}t\frac{\mathrm{d}u}{u},$

where I change the order of integration again. Now we have an inner ${t}$ integral that we can do, as it's just the standard Gaussian integral (google this if this doesn't make sense to you). The inner integral is

$\displaystyle \int_{-\infty}^\infty e^{-ut^2} \mathrm{d}t = \sqrt{\pi / u}. $

Putting this into the above yields

$\displaystyle \sqrt{\pi} \int_0^\infty u^{n-1/2} e^{-u} \frac{\mathrm{d}u}{u}, \ \ \ \ \ (4)$

which is exactly the definition for ${\Gamma(n-\frac12) \cdot \sqrt \pi}$.

But remember, we multiplied everything by ${\Gamma(n)}$ to start with. So we divide by that to get the result:

$\displaystyle \int_{-\infty}^\infty \frac{\mathrm{d}t}{(1 + t^2)^n} = \dfrac{\sqrt{\pi} \Gamma(n-\frac12)}{\Gamma(n)} \ \ \ \ \ (5)$

Finally, a copy of the latex file itself.

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Comments (1)2015-05-21 Benjamin BaerHello,

Can you elaborate on your claim that multiplying by $Gamma$ was a common integral evaluation technique in the past?

Thanks,

Ben