# Convergence test of improper integral

Is the improper integral $$\int_1^\infty \sin x\, dx$$ convergent? How about $$\int_1^\infty \sin(x^2)\,dx$$? Prove it.

So I will separate the question into two parts (a)$$\int_1^\infty \sin x\,dx$$ (b)$$\int_1^\infty \sin(x^2)\,dx$$

What would you do to answer this type of proof questions? For me, I would first check if the integral is convergent or divergent and finally give a formal proof.

For example in part a, I could get the result $$\cos1-\cos\infty$$ which is divergent.

Then I would start to provide a formal proof by comparison test.

But here is the problem. I need to set $$0\leq g(x)\leq \sin x$$ on the interval $$[1,\infty)$$. If I make $$g(x)=\sin x-1$$ then it seems to be invalid as $$1\leq \sin x\leq \sin x+1$$.

What value of g(x) should I take would be better? Also for part b I think it should also be divergent, but I have no clue how to prove it.

May be, we could use $$\sin(x^2)=\frac {e^{i x^2}-e^{-i x^2} }{2i}$$ Now, using the Gaussian integrals and the error functions, $$\int e^{a x^2}\,dx=\frac{\sqrt{\pi } }{2 \sqrt{a}}\,\text{erfi}\left(\sqrt{a} x\right)$$ $$\int_1^\infty e^{a x^2}\,dx=\frac{\sqrt{\pi } }{2 \sqrt{-a}}\,\text{erfc}\left(\sqrt{-a}\right)$$ Doing it twice and making $$a=i$$ at the end, we should get $$\int_1^\infty \sin(x^2)\,dx=\left(\frac{1}{4}-\frac{i}{4}\right) \sqrt{\frac{\pi }{2}} \left((1+i)-i\, \text{erf}\left((-1)^{1/4}\right)+\text{erf}\left((-1)^{3/4}\right)\right)$$ which must be, for sure, a finite real number.
By definition, $$\int_1^\infty \sin(x) dx=\lim_{a\to\infty} \int_1^a\sin(x)dx$$, which, as you noted, does not exist, hence the integral does not converge.
For the second integral, substitute $$u=x^2$$ so $$\frac 12 u^{-1/2} du=dx$$ and estimate via integration by parts $$\frac 12\int_1^\infty \sin(u)u^{-1/2} du=-\frac{\cos(u)}{2u^{1/2}}\Big|_1^\infty-\int_1^\infty \frac{\cos(u)}{4u^{3/2}}du$$ The first term converges because $$|\cos(u)|\le 1$$ which also implies that $$\left|\int_1^\infty \frac{\cos(u)}{u^{3/2}}du\right|\le \int_1^\infty u^{-3/2}du<\infty$$.
By changing variable, say, $$t=x^2$$, we have $$\int_1^\infty \sin(x^2) dx =\int_1^\infty\frac{\sin t}{2\sqrt{t}} dt$$ Since for any A > 1, $$|\int_1^A \sin t dt|=|\cos 1 -\cos A|\leq 2$$, and $$\frac{1}{\sqrt{t}}$$ monotone decreasing when $$t > 1$$, and $$\frac{1}{\sqrt{t}}\to 0, as$$ $$t \to\infty.$$ By Dirichlet principle, we know the improper integral converges. However, it is not an absolute convergence.