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I need your help to solve this exercise. I had to study the convergence of the integral:

$$\int_0^{\infty}{\frac{\sin(x^2+\cos(x)-1)}{x}}dx$$

I tried different methods to do it, but I got no where. Can anyone do a favor and help me to solve it?

Thanks in advance

Simply Beautiful Art
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daniel11
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  • Mathematica numerically integrates to obtain $0.7230...$ so it certainly looks possible. Also $\lim$ as $x\to 0$ and $x\to\infty$ of the integrand look good, so that's further evidence. – pshmath0 May 06 '17 at 11:11
  • does Dirichlet's test help? https://math.stackexchange.com/questions/141048/dirichlets-test-for-convergence-of-improper-integrals – Li Chun Min May 06 '17 at 11:12

1 Answers1

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Hint. The integral is convergent. Note that the integrand is integrable in $[0,1]$ and $$\int_1^{\infty}{\frac{\sin(x^2+\cos(x)-1)}{x}}dx=\int_1^{\infty}{\frac{D(-\cos(x^2+\cos(x)-1))}{x(2x-\sin(x))}}dx .$$ Then integrate by parts and show that the following integral is absolutely convergent $$\int_1^{\infty}-\cos(x^2+\cos(x)-1)D\left(\frac{1}{x(2x-\sin(x))}\right)dx$$ because $|-\cos(x^2+\cos(x)-1)|\leq 1$ and $$\lim_{x\to +\infty}\frac{D\left(\frac{1}{x(2x-\sin(x))}\right)}{1/x^2}=0.$$

Robert Z
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  • I tried to use Dirichlet's test, but there is a problem: how to prove that $\int_1^x{sin(t^2+cos(t)-1)}dt$ is bounded? – daniel11 May 06 '17 at 12:13
  • It can be done by following the same steps of my hint without the term $x$ in the denominator. In the final limit replace $1/x^2$ with $1/x^{3/2}$. – Robert Z May 06 '17 at 12:19
  • Yes. Follow the same steps of the hint. – Robert Z May 06 '17 at 13:08
  • $\int_1^x{sin(t^2+cos(t)-1)}dt=\int_{1}^{x}{\frac{d(t^2+cos(t)-1)}{2t-sin(t)}}dt=\frac{x^2+cos(x)-1}{2x-sin(x)}- \frac{cos(1)}{2-sin(1)}+\int_{1}^{x}{\frac{(2-cos(t))(t^2+cos(t)-1)}{(sin(t)-2t)^2}}dt,$ – daniel11 May 06 '17 at 13:40
  • then how to continue? – daniel11 May 06 '17 at 13:42
  • The last integral is absolutely convergent because then $|f(t)|/(1/t^{3/2})\to 0$ as $t$ goes to infinity where $f$ is the integrand (and the integral of $1/t^{3/2}$ is convergent in $[1,+\infty)$. – Robert Z May 06 '17 at 13:50
  • yes, but $lim_{x\rightarrow \infty}\frac{x^2+cos(x)-1}{2x-sin(x)}=\infty$ which mean the integral is divergent in this case – daniel11 May 06 '17 at 14:32
  • You missed the cosine. You should have $\cos(x^2+\cos(x)-1)$. – Robert Z May 06 '17 at 14:38
  • oh yes, I forget it, this means it is convergent. – daniel11 May 06 '17 at 16:18
  • can we use dirichlet's test to prove that $\int_0^{\infty}{\frac{sin(x^2+cos(x)-1)}{x}}dx$ converge? – daniel11 May 06 '17 at 16:21
  • Yes. You have just verified the hypothesis. – Robert Z May 06 '17 at 16:26
  • cause the integral is convergent, than is bounded. one more question if f and g are 2 functions and we have the integral $\int_q^{\infty}{f(x)g(x)dx}$ by integrating by parts we obtain $\int_q^{\infty}{f(x)g(x)dx}=G(x)f(x)- \int_q^{\infty}{f'(x)G(x)dx}dx$ where $G(x)=\int_q^x{g(t)dt}$. If $lim_{\infty}G(x)f(x)$ exist and $\int_q^{\infty}{f'(x)G(x)dx}dx$ converge $conditionally$, then, does $\int_q^{\infty}{f(x)g(x)dx}$ converge in this case? – daniel11 May 06 '17 at 17:08
  • Yes, It converges. – Robert Z May 06 '17 at 17:12