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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.

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3 Answers 3

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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.

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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 $\left|\int_1^\infty \frac{\cos(u)}{u^{3/2}}du\right|\le \int_1^\infty u^{-3/2}du<\infty$.

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For part(b), I think it converges.

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.

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