improper integral-integral test for convergence:$\displaystyle\int_0^{+\infty}\frac{\cos (\ln x)}{\sqrt{x^4+1}} \mathrm dx$

Question

Verify the convergence of the improper integral $$\int_0^{+\infty}\frac{\cos (\ln x)}{\sqrt{x^4+1}} \mathrm dx.$$

I split this integral into two parts. I know how to prove its convergence from $1$ to infinity, but I can't prove the convergence of the second part which is from $0$ to $1$ can someone help? I tried integrating by parts but it didn't work for me. (Sorry if the language is not fully understandable, English is not my first language the exact math terms so I searched for them on Google.)

Answer 1

You can try the absolute convergence for the $\int^1_0\frac{\cos(\ln(x))}{\sqrt{x^4+1}}dx$. With this, you can: $\int^1_0\frac{|\cos(\ln(x))|}{\sqrt{x^4+1}}dx\leq \int^1_0\frac{1}{\sqrt{x^4+1}}dx$ and value of this integral is finite, since the function is continuous and the interval is bounded. The same could be done for the $[1,+\infty)$ part.

Answer 2

$$ \begin{aligned} \int_0^{+\infty}\frac{\cos (\ln x)}{\sqrt{x^4+1}} \mathrm dx&= \int_0^1 \frac{\cos (\ln x)}{\sqrt{x^4+1}} d x+\int_1^{\infty} \frac{\cos (\ln x)}{\sqrt{x^4+1}} d x \\ & =\int_{\infty}^1 \frac{\cos \left(\ln \frac{1}{x}\right)}{\sqrt{\frac{1}{x^4}+1}} \frac{d x}{-x^2}+\int_1^{\infty} \frac{\cos (\ln x)}{\sqrt{x^4+1}} d x \quad \textrm{ via }x\mapsto \frac{1}{x} \\ & =2 \int_1^{\infty} \frac{\cos (\ln x)}{\sqrt{1+x^4}} d x \end{aligned} $$ If you had proved that $\int_1^{\infty} \frac{\cos (\ln x)}{\sqrt{1+x^4}} d x$ is convergent, then $\int_0^{+\infty}\frac{\cos (\ln x)}{\sqrt{x^4+1}} \mathrm dx$ is convergent too.

Answer 3

The integral converges absolutely by comparison with $$ \int_0^\infty \frac{dx}{\sqrt{x^4+1}} = \frac14B\left(\frac14, \frac14\right) . $$

Answer 4

Since $x^4+1\ge {1\over 2}(x^2+1)^2$ we get $${|\cos\ln x|\over \sqrt{x^4+1}}\le {\sqrt{2}\over x^2+1}$$and $$\sqrt{2}\int\limits_0^\infty {dx\over x^2+1}=\sqrt{2}\arctan x\bigg\vert _0^\infty={\pi\over \sqrt{2}} $$ Hence the integral in question is absolutely convergent.