# Divergent improper integral

I have to determine if the following integral converges/diverges. I know that the integral diverges, but I can't find a way to prove it.

$$\displaystyle\int_1^{+\infty}\frac{1-\cos(x)}{\left(\sqrt{1+x^2}-1\right)\arctan\left(\sqrt{x}\right)}\,\mathrm dx$$.

Since

$$\left(\sqrt{1+x^2}-1\right)\arctan(\sqrt{x})\underset{x\to+\infty}{\sim}x\dfrac{\pi}{2}$$

I can write the integral as

$$\displaystyle\int_1^{+\infty}\frac{1-\cos(x)}{x\frac{\pi}{2}}\,\mathrm dx=\frac{2}{\pi}\int_1^{+\infty}\frac{1-\cos(x)}{x}\,\mathrm dx$$

I'm trying to find a function $$\,0\leqslant g(x)\,$$ such that $$\displaystyle\,\int_1^{+\infty}\!\!\!\!\!g(x)\,\mathrm dx\,$$ is divergent and $$\displaystyle\,\int_1^{+\infty}\!\!\!\!\!g(x)\,\mathrm dx\leqslant\int_1^{+\infty}\!\frac{1-\cos(x)}{x}\,\mathrm dx\;,$$

or $$\;g(x)\underset{x\to+\infty}{\sim}\dfrac{1-\cos(x)}{x}\,$$

but I'm really struggling because of that cosine.

Any advice would be greatly appreciated!

By applying integration by parts, we get that

$$\displaystyle\int_1^{+\infty}\!\frac{1-\cos x}x\mathrm dx=\left[\dfrac{x-\sin x}x\right]_1^{+\infty}\!\!+\int_1^{+\infty}\!\frac{x-\sin x}{x^2}\mathrm dx=$$

$$\displaystyle=\sin 1+\int_1^{+\infty}\!\frac{\mathrm dx}x-\int_1^{+\infty}\!\dfrac{\sin x}{x^2}\mathrm dx=+\infty\;\;,$$

indeed ,

$$\displaystyle\int_1^{+\infty}\frac{\mathrm dx}x=\big[\ln x\big]_1^{+\infty}=+\infty\quad$$ and

$$\displaystyle\left|\int_1^{+\infty}\!\frac{\sin x}{x^2}\mathrm dx\right|\leqslant \int_1^{+\infty}\!\frac{\left|\sin x\right|}{x^2}\mathrm dx\leqslant\,\int_1^{+\infty}\!\frac{\mathrm dx}{x^2}=1\,.$$

• Thank you so much!
– rik
Jun 14 at 21:32
• @rik, you are welcome ! Jun 14 at 21:34