# Infinite Integral of Trigonometric Functions

I am interested in finding the Integral: $I = \int\limits_{0}^{\infty} \sin x \,dx$. Clearly going the conventional way $I = -\cos (\infty) + \cos(0)$ will not lead to a definite answer. However I have thought of the problem in a different way. First, we already know from Fourier Transform that $\int\limits_{-\infty}^{\infty}\exp \left(i \omega t\right) \, dt = 2 \pi \delta\left(\omega \right)$. Thus , upon setting $\omega = 1$, we have $\int\limits_{-\infty}^{\infty}\cos \left(t\right) \, dt = 0$. But knowing that $\cos(t)$ is even in $t$ thus, $0 = \int\limits_{-\infty}^{\infty}\cos \left(t\right) \, dt = 2 \int\limits_{0}^{\infty}\cos \left(t\right) \, dt \Rightarrow \int\limits_{0}^{\infty}\cos \left(t\right) \, dt = 0$.

Now consider the transformation $u = t + \frac{\pi}{2}$, we thus have $0 = \int\limits_{\frac{\pi}{2}}^{\infty}\cos \left(u - \frac{\pi}{2}\right) \, du = \int\limits_{\frac{\pi}{2}}^{\infty}\sin \left(u \right) \, du = \int\limits_{\frac{\pi}{2}}^{0}\sin \left(u \right) \, du + \int\limits_{0}^{\infty}\sin \left(u \right) \, du = \int\limits_{0}^{\infty}\sin \left(u \right) \, du -1 \Rightarrow \int\limits_{0}^{\infty}\sin \left(u \right) \, du = 1$.

This solution is even sustained by the Laplace Transform results where we have $\mathcal{L} \lbrace \sin\left( t\right)\rbrace = \dfrac{1}{s^2 + 1}$. But $\mathcal{L} \lbrace \sin\left( t\right)\rbrace = \int\limits_{0}^{\infty}\exp \left(-st \right) \sin \left(t \right) \, dt$

Thus $\int\limits_{0}^{\infty}\sin \left(t \right) \, dt = \lim\limits_{s \rightarrow 0} \int\limits_{0}^{\infty}\exp \left(-st \right) \sin \left(t \right) \, dt = \lim\limits_{s \rightarrow 0}\dfrac{1}{s^2 + 1} = 1$. Hence $\int\limits_{0}^{\infty}\sin \left(t \right) \, dt = 1$.

I am here getting exactly the same result for the integral by computing it in two different methods. Is my work correct? I definitely know that $\sin$ is not Reimann-integrable over the interval $[0,\infty)$. Is there a special name for this integral, or its evaluation in this method?

Thanks very much for your suggestions!

• I have heard this technique (through the Laplace/Mellin transform) called "Zeta regularization", but honestly I do not know if it is the standard name. Jul 9, 2015 at 13:35
• One may also argue that $$\int_{0}^{+\infty}x^p \sin(x)\,dx = \Gamma(p+1)\cos\frac{\pi p}{2}$$ for any $p$ such that $-2<\text{Re}(p)<0$, also leading to $1$ as regularized value. Jul 9, 2015 at 13:45
• You are entering the world of Generalized Functions or Distributions. One of the most notable treatises on this topic, the seminal one, is HERE. Gel'fand and Shilov rock! Jul 9, 2015 at 14:36

One way to a sensible regularization is to interpret the integral as $$I=\int_0^{\infty}\sin(x)\underbrace{=}_{def.}\lim_{\delta \rightarrow 0_+}\int_0^{\infty}\sin(x)e^{-\delta x}$$

Now performing the trival integration we have $$I=\lim_{\delta \rightarrow 0_+}\frac{1}{1+\delta^2}=1$$

as suggested by your regularization attempts

Another regularization procedure could be introduced by

$$\tilde{I}=\int_0^{\infty}\sin(x)\underbrace{=}_{def.}\lim_{\delta \rightarrow 0_+}\int_0^{\infty}\sin(x)e^{-\delta x^2}$$

This can be easily integrated in terms of error functions: $$\tilde{I}=\lim_{\delta \rightarrow 0_+}\frac{e^{-\frac{4}{\delta}}\text{Erfi}[\frac{1}{2\sqrt{\delta}}]}{2\sqrt{\delta}}=\lim_{\delta \rightarrow 0_+}\frac{F\left(\frac{1}{2 \sqrt{\delta }}\right)}{\sqrt{\delta}}$$

where $\text{Erfi}[x]$ is error function of imaginary argument and $F(x)$ is the Dawson integral which for $x\rightarrow \infty$ behaves as $F(x)\sim \frac{1}{2x}$ and therfore

$$\tilde{I}=1=I$$

so both regularisation procedures are consistent

• I am no expert on advanced analysis, but this method of regularisation seems very natural to me, especially from a Physics perspective. One can think of this as the limit of idealised underdamped harmonic oscillation where the damping coefficient tends to zero. But I can't quite make the full connection for the integral to the physical analogy. Jul 9, 2015 at 14:15

Hint: Use Euler's formula in conjunction with $~\displaystyle\int_0^\infty e^{-kx}~dx=\frac1k~$ for $~\Re(k)>0$.

There are lots of related techniques for assigning finite values to divergent sums and integrals, and they often agree on a value. You should look into the terms Cauchy principle value, Cesàro summation, or the Abel sense of convergence. Other regularizations work by introducing an artificial parameter, and then taking the limit as the parameter vanishes. This often works for physical problems where concepts like "loss" are vanishingly small but cannot by zero by the laws of thermodynamics. I've often seen these vanishing loss approaches also agree with the above divergent sum/integral techniques.

As for whether this is "right": you can come up with other schemes that give other, different finite values. Consider: $$\int_0^\infty \sin(x)\;dx = \sum_{n=0}^\infty \int_{2\pi n}^{2\pi(n+1)} \sin(x)\;dx = 0+0+0\ldots = 0$$ In fact, I'm confident that you could convince yourself that the value of that integral is any value, depending on how you sum things up. Fourier and Laplace techniques are rooted in practical physical calculations however, so when these kinds of questions arise in approximations to the real world, those often give results that are consistent with observations. However, this does not make the mathematics correct or rigorous.

I will try to be as elementary as possible :

Consider the following sum :

$$S=\sum_{n=0}^\infty (-1)^n$$

This sum has value $$1/2$$ as if we shift the sum by one term and if we add it to itself then we get $$2S=1$$

Now if we consider the graph of $$sinx$$ it's periodic with period 2π and from $$0$$ to $$π$$ it's integration is 2 and from $$π$$ to $$2π$$ it's$$-2$$ due to symmetry around$$π$$.

So integral in your question reduces to $$2S$$ which is 1

I am interested in finding the Integral: $$I = \int\limits_{0}^{\infty} \sin x \,dx$$

This is simple. Following Cesaro integration,

$$\int\limits_{0}^{\infty} \sin x \,dx=1$$

and

$$\int\limits_{0}^{\infty} \cos x \,dx=0$$

In other words, you have to subtract the value of negative cosine at zero (-1) from the mean value of negative cosine at infinity (0). You get 1.