# $\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$ similar identities

I recently found that $$\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}=\pi$$ iff $$b^2-4ac=-4$$ I found it by integrating $$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{ax^2+bx+c}$$ If the reciprocal of the function is to be integrated over the entire real line, then the function must not have any real zeros. This implies that $$b^2-4ac<0$$ With this in mind, we complete the square: $$I=\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}}$$ Then setting $$g=c-\frac{b^2}{4a}$$, and $$x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$$, $$I=\sqrt{\frac{g}{a}}\int_{-\pi/2}^{\pi/2}\frac{\sec^2u\ \mathrm{d}u}{g\tan^2u+g}=\frac{\pi}{\sqrt{ag}}$$ So our identity holds for $$ag=1$$ And with a little algebra, $$b^2-4ac=-4$$

So my question is, are there any other similar identities involving other famous constants? Cheers!

• What restrictions (if any) exist on $a,b,c$ ? are they natural numbers? rational numbers? real? complex? etc? – user150203 Dec 15 '18 at 9:22
• I also decided to plot the relationship between a,b,c on wolphramalpha and found out that the relationship defined is actually a 'Two-Sheeted Hyperboloid' mathworld.wolfram.com/Two-SheetedHyperboloid.html – user150203 Dec 15 '18 at 10:19
• @DavidG I would think they are real, but I'm not sure... – clathratus Dec 15 '18 at 16:59
• Related (not in the sense this is a duplicate, but it may interest the same people this does): math.stackexchange.com/questions/2821112/integral-milking – J.G. Dec 15 '18 at 18:00

First if we consider the integral:

$$$$I(a) = \int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx = \frac{\pi}{a}$$$$

Then we can see:

$$$$I(a) + cI(b) = \left[\int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx + c\int_{-\infty}^{\infty}\frac{1}{x^2 + b^2}\:dx \right] = \frac{\pi}{a} + c\frac{\pi}{b} = \pi\left[\frac{1}{a} + c\frac{1}{b} \right]$$$$

We can see that this can be expanded to any number of $$I(a)$$ terms and any sequence of $$c_i$$ values, i.e.

$$$$\sum_{i = 1}^{\infty} (-1)^{i + 1} \frac{1}{i + 1}I(i) = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)$$$$

Thus,

$$$$\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)$$$$

Or as another example,

$$$$\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{1}{x^2 + i^4} \:dx = \frac{ \pi^3}{6}$$$$

What's even better (IMO) is that we can take each integral spoken to and apply (through Glasser Master Theorem)

$$x = t - \sum_{i = 1}^{n - 1}\frac{\left| d_{i}\right|}{ t - e_{i}}$$

Where $$d_i, e_i \in \mathbb{R}$$ and $$n \in N$$ that the value of the integrals remain unchanged!!

As an addendum to the example:

$$$$\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)$$$$

If we call on this integral as addressed here we see that

$$$$2\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = 2\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx =\pi\ln(2)$$$$

Or, $$$$\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \frac{\pi\ln(2)}{2}$$$$

• Woah! that's really cool! Thanks David (+1) – clathratus Dec 15 '18 at 17:03

I'm not sure if this example's interesting enough to be an answer, because it can get any result you want with different linearly independent choices of the integrand, so if pressed I'll convert it into a comment.

As has been noted here before, any $$f$$ analytic on the unit disk in $$\Bbb C$$ satisfies $$\int_0^{2\pi}\Re f(e^{i\phi})d\phi=2\pi\Re f(0),$$which can be used to get a daunting-looking integral expression for any real value. I've designed some deliberately not-too-complicated examples (so let's hope I made no mistakes in my mental arithmetic).

To ensure that integral is equal to some $$k>0$$ choose$$f(z):=\exp\frac{\ln\frac{k}{2\pi}}{1+z/2},$$and multiply this choice of $$f$$ by $$z$$ or $$z-1$$ to get $$-k$$ instead. If you try this as an exercise, you'll find for $$k\ne 1$$ the real integral you obtain is pretty gnarly (though nowhere near as much as it could be with some other choice of $$f$$), which is deliberate. If you want a far-from-obvious expression for $$1$$ that follows the same way, try $$f(z)=\exp\frac{z-2\ln 2\pi}{z+2}$$instead.

• I can't wait to learn complex analysis so I can fully appreciate how cool this is. Thanks! – clathratus Dec 15 '18 at 19:17