How to use the comparison test for this integral?

Question

$$\int_{π}^{∞} \frac{\sin(x)}{\sqrt{x} + \sin(x)} dx .$$

Need to prove the integral diverges. Tried to use all the different techniques but to no avail.

Answer 1

you can see the integral as the sum integral between n^2pi and (n+1)^2pi the dominator in this interval is bigger than 1/((n+1)sq root of pi)+1) so the integral become the sum k*1/((n+1)sq root of pi)+1) where k is the integral of sin between n^2pi.... and that series diverges I wish I can write with programation lg to be more helpful

Answer 2

Using the Taylor expansion $1/(1+\varepsilon)=1-\varepsilon+O(\varepsilon^2) $, $$ \frac{\sin(x)}{\sqrt{x} + \sin(x)}=\frac{\sin(x)/\sqrt{x}}{1 + \sin(x)/\sqrt{x}}=\sin(x)/\sqrt{x}\left(1-\sin(x)/\sqrt{x}+O(1/x)\right)=\\\sin(x)/\sqrt{x}-\sin^2(x)/x+O(x^{-3/2}) $$ Now $\int_\pi^\infty \sin(x)/\sqrt{x}\; dx$ is convergent since the series $\sum_n (-1)^n/\sqrt{n} $ is convergent (alternating signs and terms tending to zero). In fact, write $$ \int_{π}^{∞} \sin(x)/\sqrt{x}\; dx = \sum_n \int_{nπ}^{(n+1)\pi} \sin(x)/\sqrt{x}\; dx $$ and you easily see that the integral and the series $\sum_n (-1)^n/\sqrt{n} $ are comparable. Similarly, $\int_\pi^\infty \sin^2(x)/x\; dx$ is divergent since the series $\sum_n 1/n $ is divergent. For this, observe that: $$ \int_{nπ}^{(n+1)\pi} \frac{\sin^2(x)}{x}\; dx\ge \int_{nπ}^{(n+1)\pi} \frac{\sin^2(x)}{(n+1)\pi}\; dx=\frac{1}{2 (n+1)}. $$

Of course $\int_\pi^\infty O(x^{-3/2}) \; dx$ is convergent.

So you have two convergent contributions and a divergent one, hence the integral is divergent (and negative)