Determining the convergence of $\int_0^{1/2} \frac{\cos(x)}{\sqrt x + x} dx$

Question

Determine the convergence of the integral $$\int_0^{1/2} \frac{\cos(x)}{\sqrt x + x} dx$$

I have tried everything to try test for convergence. I have tried to integrate it with no success by using parts and substitution. Each time I apply integration by parts it becomes more complex with each computation.

Any other suggestions besides trying to evaluate it and then plugging in the limits?

Thanks.

Answer 1

Find an equivalent function near $0$ by stating an equivalent form for $\cos x$ and $\frac{1}{\sqrt{x} + x}$ :

Near $0$, you have \begin{align} \cos x &= 1 + o(x) \\ \frac{1}{\sqrt{x} + x} &= \frac{1}{x(1 + \sqrt{1/x})} = 1/x(1 - \frac{1}{\sqrt{x}} + o(1/\sqrt{x})) \end{align} And now you have \begin{align} f(x) = 1/x + o(something) \end{align} if "something" = o(1/x), you have $f(x) \sim 1/x$ which is positive and non-integrable at $0$, thus $f$ is not integrable at $0$

Edit : sorry, i did a huge mistake. My mistake is that the second development is not true. Te same idea replacing $1/x$ and $1/\sqrt{x}$ show that the fraction is equivalent to $1/\sqrt{x}$ at $0$ which is positive and integrable

Answer 2

For convergence, the problem being at the lower bound, use Taylor series to get $$\frac{\cos(x)}{\sqrt x + x}=\frac{1}{\sqrt{x}}-1+\sqrt{x}-x+O\left(x^{3/2}\right)$$ So, no problem.

Use more terms and integrate termwise for a quite good approximation.

Edit

Using this beautiful approximation ($\color{red}{1,400}$ years old !!) $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$

$$2 \left(1+\pi ^2\right) \int_0^{1/2} \frac{\cos(x)}{\sqrt x + x} dx=$$ $$-16 \log \left(1+\frac{1}{\sqrt{2}}\right)+\pi ^2 \left(-5+4 \log \left(1+\frac{1}{\sqrt{2}}\right)+10 \log (\pi )-5 \log \left(\frac{1}{4}+\pi ^2\right)\right)+10 \pi \tan ^{-1}(2 \pi )-5 \sqrt{2} \pi ^{3/2} \left(\tan ^{-1}\left(\frac{2 \sqrt{\pi }}{1-2 \pi }\right)+\tanh ^{-1}\left(\frac{2 \sqrt{\pi }}{1+2 \pi }\right)\right)+5 \sqrt{2 \pi } \left(\tan ^{-1}\left(\frac{2 \sqrt{\pi }}{2 \pi -1}\right)+\tanh ^{-1}\left(\frac{2 \sqrt{\pi }}{1+2 \pi }\right)\right)$$ making the approximation $$ \int_0^{1/2} \frac{\cos(x)}{\sqrt x + x} dx\approx 1.04727 $$ while numerical integration gives $1.04751$.

Answer 3

$f(x) = \frac{\cos(x)}{\sqrt{x}+x} \sim \frac{1}{\sqrt{x}}$ as $x\to0^+$, $f(x)\geq 0$ near $0$ and $\int_0^{1/2}\frac{1}{\sqrt{x}} dx $ converges.

Therefore, $\int_0^{1/2} f(x) dx $ converges.

Answer 4

We have $$\int_0^{1/2}\frac{\cos{x}}{\sqrt x+x}\,\mathrm{d}x\leq \int_0^{1/2}\frac{\mathrm{d}x}{\sqrt x+x}=\\ \int_0^{\sqrt{1/2}}\frac{2\mathrm{d}u}{1+u}= 2\ln\left(1+\frac1{\sqrt2}\right) $$