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 $$\lim_{\epsilon\to 0}I_n=i\pi (n+1)!. \quad \quad \quad \quad (10) $$