What is the $\varepsilon\to 0$ limit of the following double integral
$$\int\limits\_{-1}^1d\tau\;\sqrt{1-\tau^2}\;\tau\int\limits\_0^\infty dq\;q^2e^{iq(\tau+i\varepsilon)}\;?$$
I was asked about this integral by my friend who got it in a physics research project. In fact this is just $n=1$ case of a more general integral
$$\int\limits\_{-1}^1d\tau\;(1-\tau^2)^{n/2}\;\tau\int\limits\_0^\infty dq\;q^{n+1}e^{iq(\tau+i\varepsilon)}.$$
For $n$ even, the integral can be calculated rather simply by using
$$\int\limits\_0^\infty dq \, e^{iq(\tau + i\varepsilon)} \,=\, 
\frac{\varepsilon}{\tau^2 \,+\, \varepsilon^2} \,+\, \frac{i\tau}
{\tau^2 \,+\, \varepsilon^2} \,\to\, \pi\delta(\tau) \,+\, 
\mathcal{P}\frac{i}{\tau},$$ because in this case only the delta-function contributes (or more precisely its derivatives, as far as the original integral is concerned). But for $n$ odd, only the principal value part contributes and things become messy. 



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I believe Igor Khavkine's method of using Hadamard regularization is indeed most suitable for this integral. By the way, an useful reference about Hadamard principal value is <http://cms.math.ca/10.4153/CJM-1957-015-1> (A generalization of the Cauchy principal value, by Charles Fox, Canad. J. Math. 9(1957), 110-117). However, brute force $\epsilon$-regularization method also works, as described below.


We can transform the initial integral in the following way
$$I\_n=\int\limits\_{-1}^1d\tau\;(1-\tau^2)^{n/2}\tau\int\limits\_0
^\infty q^{n+1}e^{iq(\tau+i\epsilon)}dq=$$ $$ (-1)^{n+1}\frac{d^{n+1}}
{d\epsilon^{n+1}}\int\limits\_{-1}^1d\tau\;(1-\tau^2)^{n/2}\tau\int\limits\_0
^\infty e^{iq(\tau+i\epsilon)}dq=$$ $$
i(-1)^{n+1}\frac{d^{n+1}}{d\epsilon^{n+1}}\int\limits\_{-1}^1d\tau\;
(1-\tau^2)^{n/2}\frac{\tau^2+\epsilon^2-\epsilon^2}{\tau^2+\epsilon^2}=$$
$$i(-1)^n\frac{d^{n+1}}{d\epsilon^{n+1}}\left[\epsilon^2\int\limits\_{-1}^1
\frac{(1-\tau^2)^{n/2}}{\tau^2+\epsilon^2}\;d\tau\right].$$
Let us make the substitution $\tau=\sin{\theta}$, then
$$I\_n=i(-1)^n\frac{d^{n+1}}{d\epsilon^{n+1}}\left[
\epsilon^2\int\limits\_{-\pi/2}^{\pi/2}\frac{\cos^{n+1}{\theta}}
{\sin^2{\theta}+\epsilon^2}\;d\theta\right]\equiv
i(-1)^n\frac{d^{n+1}}{d\epsilon^{n+1}}\left[\epsilon^2J\_{n+1}(\epsilon)
\right]. \;\;\;\; (1)$$
We have
$$J\_n(\epsilon)=\int\limits\_{-\pi/2}^{\pi/2}\frac{\cos^n{\theta}}
{\sin^2{\theta}+\epsilon^2}\;d\theta=\int\limits\_{-\pi/2}^{\pi/2}
\frac{\cos^{n-2}{\theta}(1-\sin^2{\theta}-\epsilon^2+\epsilon^2)}
{\sin^2{\theta}+\epsilon^2}\;d\theta,$$
and hence
$$J\_n(\epsilon)=(1+\epsilon^2)J\_{n-2}(\epsilon)-K\_{n-2},\quad \quad \quad \quad (2)$$
where
$$K\_n=\int\limits\_{-\pi/2}^{\pi/2}\cos^n{\theta}\;d\theta.$$
Using (2), we can prove by induction
$$J\_{2m}=(1+\epsilon^2)^mJ\_0-\sum\limits\_{i=0}^{m-1}(1+\epsilon^2)^i
K\_{2(m-1)-2i}, \quad \quad \quad \quad (3) $$
$$J\_{2m+1}=(1+\epsilon^2)^mJ\_1-\sum\limits\_{i=0}^{m-1}(1+\epsilon^2)^i
K\_{2m-1-2i}. \quad \quad \quad \quad (4)$$
Therefore, if $n=2m-1$ is odd, we get from (1) and (3)
$$I\_{2m-1}=-i\frac{d^{2m}}{d\epsilon^{2m}}\left
[\epsilon^2(1+\epsilon^2)^mJ\_0(\epsilon)-\epsilon^2(1+\epsilon^2)^{m-1}
K\_0\right ], \quad \quad \quad \quad (5)$$
all other terms from (3) giving zero. But
$$J\_0(\epsilon)=\int\limits\_{-\pi/2}^{\pi/2}\frac{d\theta}{\sin^2{\theta}+
\epsilon^2}=\frac{\pi}{\epsilon\sqrt{1+\epsilon^2}},$$
and $K\_0=\pi$. Therefore (5) reduces to
$$I\_{2m-1}=(2m)!i\pi-i\pi\frac{d^{2m}}{d\epsilon^{2m}}\left[\epsilon
(1+\epsilon^2)^{m-1/2}\right ],$$
which is the same as
$$I\_{2m-1}=(2m)!i\pi-\frac{i\pi}{2m+1}\frac{d^{2m+1}}{d\epsilon^{2m+1}}
(1+\epsilon^2)^{m+1/2}. \quad \quad \quad \quad (6)$$
In the $\epsilon\to 0$ limit, the second term in (6) gives zero,
because the Taylor expansion of $(1+\epsilon^2)^{m+1/2}$ for small
$\epsilon$ contains only even powers of $\epsilon$. Therefore, finally
$$\lim\_{\epsilon\to 0}I\_{2m-1}=(2m)!i\pi. \quad \quad \quad \quad (7)$$
If $n=2m$ is even, we get similarly
$$I\_{2m}=i\frac{d^{2m+1}}{d\epsilon^{2m+1}}\left[\epsilon^2(1+\epsilon^2)^m
J\_1(\epsilon)\right],$$
because no term containing $K$-factors survives after taking $(2m+1)$-th
derivative. Using
$$\int\limits\_{-\pi/2}^{\pi/2}\frac{\cos{\theta}}{\sin^2{\theta}+\epsilon^2}
\;d\theta=\frac{2}{\epsilon}\arctan{\left(\frac{1}{\epsilon}\right)},$$
we get
$$I\_{2m}=2i\frac{d^{2m+1}}{d\epsilon^{2m+1}}\left [\epsilon(1+\epsilon^2)^m
\arctan{\left(\frac{1}{\epsilon}\right)}\right ]. \quad \quad \quad \quad (8) $$
In the $\epsilon\to 0$ limit, since $\arctan{\left(1/\epsilon\right)}$ and its
derivatives are not singular in this limit, the application of the Leibniz
rule to (8) gives
$$\lim\_{\epsilon\to 0}I\_{2m}=2i\lim\_{\epsilon\to 0}
\arctan{\left(\frac{1}{\epsilon}\right)}\frac{d^{2m+1}}{d\epsilon^{2m+1}}
[\epsilon(1+\epsilon^2)^m]=i\pi(2m+1)! \quad \quad \quad \quad (9) $$
All other terms vanish, because, for any $i\ge 1$, either
$$\lim\_{\epsilon\to 0}\frac{d^{2i}}{d\epsilon^{2i}}
[\epsilon(1+\epsilon^2)^m]=0,$$
since binomial expansion of $\epsilon(1+\epsilon^2)^m$
contains only odd powers of $\epsilon$, or
$$\lim\_{\epsilon\to 0}\frac{d^{2i}}{d\epsilon^{2i}}
\arctan{\left(\frac{1}{\epsilon}\right)}=-\lim\_{\epsilon\to
0}\frac{d^{2i-1}}{d\epsilon^{2i-1}}\frac{1}{1+\epsilon^2}=0.$$
Equations (7) and (9) can be unified in the final result
$$\lim\_{\epsilon\to 0}I\_n=i\pi (n+1)!. \quad \quad \quad \quad (10) $$



Limit of a double integral
I believe these integrals can be evaluated directly in the $\varepsilon=0$ limit by interpreting the result of the inner integral as a distribution.
$$
\int\limits\_0^\infty dq \, q^{n+1} \, e^{iq(\tau + i\varepsilon)}
 \to (-i)^{n+1}\frac{\partial^{n+1}}{\partial\tau^{n+1}} \left(\pi\delta(\tau) + \mathcal{P}\frac{i}{\tau}\right).
$$
The rest of the integrand is $\sqrt{1-\tau^2}\tau$, which is odd, so that for odd $n$ only the second term contributes. Evaluating the derivatives gives this term as
$$
i(-i)^{n+1}\frac{\partial^{n+1}}{\partial\tau^{n+1}}\mathcal{P}\frac{1}{\tau} =
 i(-i)^{n+1}\frac{\partial^{n+2}}{\partial\tau^{n+2}}\log|\tau| =
 -i^n (n+1)! \mathcal{P} \frac{1}{\tau^{n+2}},
$$
where the $\mathcal{P}$ now stands for *finite part* rather than *principal value*. This distribution can be defined simply as the distributional derivative as in the middle expression or it can be defined using the [Hadamard regularization](http://en.wikipedia.org/wiki/Hadamard_finite_part), which consists of introducing an integration cutoff (say $\sigma$) at $\tau=0$ and taking $\sigma\to0$ after subtracting the unique polynomial in $1/\sigma$ that makes the limit finite (presuming no $\log|\sigma|$ terms are present).


The original integral is
\begin{align\*}
I(n)
 &= -i^n(n+1)!\int\_{-1}^1 d\tau \, \mathcal{P} \frac{(1-\tau^2)^{n/2}}{\tau^{n+1}} \\
 &= -2^{-n}i^n(n+1)! \int\_{-1}^1 dt \, \mathcal{P} \frac{1}{t^{n+1}}
 \frac{(1-t^2)^{n+1}}{(1+t^2)} \\
 &= 2^{-n}i(-1)^{\frac{n+1}{2}}(n+1)! \, J(n),
\end{align\*}
where I've made the substitution $\tau=\frac{2t}{1+t^2}$, which rendered the integrand rational.


Now, the Hadamard regularization can be applied to evaluate the $J(n)$ integral for any $n$: $J(1)=-2\pi$, $J(3)=8\pi$, $J(5)=-32\pi$, ... I'll extrapolate this to $J(n) = (-1)^{\frac{n+1}{2}} 2^n \pi$, which gives the final answer $$I(n) = i\pi (n+1)! \, .$$


Incidentally, to implement the Hadamard regularization, it suffices to apply the "student integration rule": find an anti-derivative and evaluate the difference at the end-points. With this trick, the extrapolation for $J(n)$ can be justified by computing the generating function for $J(n)$:
\begin{align}
 j(z) &= \sum\_{n=0}^\infty z^n J(n-1)
 = \int\_{-1}^1 dt\, \mathcal{P} \frac{1}{1+t^2} \frac{1}{1-z (1-t^2)/t} \\
 &= \frac{\pi}{2(1+4z^2)} - \frac{\log(-1)z}{4z^2+1} ,
\end{align}
where the integral was evaluated using computer algebra. The annoyingly vague $\log(-1)$ term actually doesn't matter, since for odd values of $n$, we only need the $z$-even part of the generating function:
$$
 J(-1) + z^2 J(1) + z^4 J(3) + \cdots
 = \frac{\pi}{2(1+4z^2)}
 = \sum\_{n=-1,1,3,5,\ldots} z^{n+1} (-1)^{\frac{n+1}{2}} 2^n \pi .
$$


