Proving that this integral exists both as Riemann Integral and Lebesgue integral Q 10.9  Apostol Mathematical analysis
10.9 (a) If p>1, prove that the integral $\int_{1}^{+\infty} x^{-p} sin x  dx$ exists  both as improper riemann integral and lebesgue integral.
For proving riemann integral I have to prove that for all b$\geq$ a Lim $ b\to \infty \int_{a}^{b} f(x) dx$ exists and for proving lebesgue integral I need to prove that there exists a positive integer M such that $\int_{b}^{a}|f(x)| d x \leq M$ for every b$\geq a$.
But I am not able to prove any if the results.
For !st I used integration by parts 1 time and in integral in 2nd term  I got power as $1/x^{p+1} $ , so I stopped as power would further increase .
But If i  use improper integral test with $1/x^{1+\delta}$ ,$\delta >0$ we will get that integral is convergent as $\int_{1}^{\infty}1/x^{1+\delta} dx$  .
But I am not sure how to show the condition for existence of lebesgue integral.
Can you please help with that
Edit: I have a question suppose that $\int_{1}^{\infty}x^{-p} sin x dx $ and 0<p$\leq 1$ then How to prove that The integral exists as improper integral . The method used above can't be usedas p<1 so I am not able to proceed with proving why it is reimann integrable.
Ty
 A: Using integration by parts is a fine approach and, in fact, shows that the improper Riemann integral converges for $p > 0$.  This is also a consequence of the Dirichlet test.
The question only considers $p > 1$, and showing that the improper Riemann integral  converges in this case is even easier.   We have  $|x^{-p} \sin x | \leqslant x^{-p}$ and $\int_1^\infty x^{-p} \, dx = = 1/(p-1) < \infty$ when $p > 1$. Hence, the improper Riemann integral converges as a consequence of the Weierstrass M-test.
For Lebesgue integrability, with $f(x) = x^{-p} \sin x$ and since the Riemann and Lebesgue integrals coincide on finite intervals, we have
$$\int_{[1,\infty]}|f|\, \chi_{[1,b]}= \int_{[1,b]}|f| = \int_1^b |f(x)| \, dx $$
Since $f$ is nonnegative, it follows from the monotone convergence theorem and the convergence of the improper integral of $|f|$ that
$$\int_{[1,\infty]}|f|= \int_{[1,\infty]}\lim_{b \to \infty}|f|\, \chi_{[1,b]}= \lim_{b \to \infty} \int_{[1,\infty]}|f|\, \chi_{[1,b]} = \lim_{b \to \infty}\int_1^b |f(x)| \, dx < \infty$$
