In what sense is $\int_{-\infty }^{\infty } \frac{x}{x^2+1} \, dx = \pi i$?

Question

Suppose we want to give a meaning to the divergent integral $$I = \int_{-\infty }^{\infty } \frac{x}{x^2+1} \, dx,$$ perhaps in the sense of distributions or something (similarly to how $\int_{-\infty }^{\infty } x e^{i k x} dx$ can be given meaning as the derivative of the delta function).

Because the integrand is odd, it seems sensible to say $I = 0$. But for all $a$ for which the integral converges, $$I = \int_{-\infty }^{\infty } \frac{x^{a-1}}{x^2+1} \, dx = \frac{1}{2} \pi \left( 1 -(-1)^a\right) \csc \left(\frac{\pi a}{2}\right).$$

Analytically continuing this to $a = 2$ gives $I = \pi i$. Added: $$\lim \limits_{\epsilon \rightarrow 0} \int_{-\infty }^{\infty } \frac{x \, e^{i \epsilon x}}{x^2+1}dx = \pi i.$$ I cannot justify switching integral and limit...

Is there a unique and 'useful' way to give a meaning to $I$ in some mathematically rigorous sense?

Answer 1

By the $\Gamma$ reflection formula it follows that: $$I(a)=\int_{-\infty}^{+\infty}\frac{x^{a-1}}{x^2+1}\,dx = \pi\sin\frac{\pi a}{2}$$ for any $a$ in the strip $0<\Re(a)<2$, so the analytic continuation gives $I(2)=0$, as expected.

Answer 2

More playing around.

If any of the steps I do here are valid, please tell me so I can remove them.

$I = \int_{-\infty }^{\infty } \frac{x}{x^2+1} \, dx,$

$\begin{array}\\ \frac{a(x-i)+b(x+i)}{(x+i)(x-i)} &=\frac{(a+b)x+(b-a)i}{(x+i)(x-i)}\\ &=\frac{x}{(x+i)(x-i)} \text{ if } (b-a)=0, b+a=1 \text{, or } a=1/2, b=1/2\\ so\ \frac{x}{x^2+1}&=\frac{1/2}{x+i}+\frac{1/2}{x-i}\\ \end{array} $

Now watch closely.

$\begin{array}\\ \int_{-\infty }^{\infty } \frac{1}{x+i} \, dx &=\mid_{-\infty }^{\infty } \ln(x+i)\\ &=\lim_{x \to \infty}(\ln(x+i)-\ln(-x-i))\\ &=\lim_{x \to \infty}\ln\frac{x+i}{-x-i}\\ &=\lim_{x \to \infty}\ln\frac{1+i/x}{-1-i/x}\\ &= \ln(-1)\\ &=\pi i \end{array} $

Similarly, $\int_{-\infty }^{\infty } \frac{1}{x-i} \, dx =\pi i $.

Therefore

$\begin{array}\\ \int_{-\infty }^{\infty } \frac{x}{x^2+1} \, dx &=\int_{-\infty }^{\infty } \left(\frac{1/2}{x+i}+\frac{1/2}{x-i}\right) \, dx\\ &=\int_{-\infty }^{\infty } \frac{1/2}{x+i}\, dx+\int_{-\infty }^{\infty }\frac{1/2}{x-i}\, dx\\ &= \frac{\pi i}{2}+\frac{\pi i}{2}\\ &=\pi i \end{array} $

I'll bet Euler would agree.