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!

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

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.