Riemann-integrable, Lebesgue-integrable

Hello, I have to decide whether $$\int_{\mathbb{R}}\frac{x}{\sqrt{1+x^4}}\, \mathrm dx$$ is (a) Improper Riemann-integrable, or (b) Lebesgue-integrable.

I search for an improper Riemann-integrable majorant for $$f(x):=\frac{x}{\sqrt{1+x^4}}$$, and a non-Riemann-integrable minorant for $$f(x)$$...

Majorant:

$$\frac{\lvert x\rvert}{\sqrt{1+x^4}}\leq \ldots$$

But I don't know how to estimate the denominator...

• Hint: $1+x > x$. Commented Sep 25, 2013 at 16:21

Hint: $$\frac{x}{\sqrt{1+x^4}}\sim \frac1x, \quad x \to \infty.$$

For $x>1$ we have $x^4<1+x^4<2x^4$. Then

$$\frac{1}{\sqrt{2}\,x}< \frac{x}{\sqrt{1+x^4}} < \frac{1}{x} \;\;\mbox{ if } x>1.$$

• I have for the majorant: $\lvert f(x)\rvert\leq\frac{1}{\lvert x\rvert}$. But is $\int_{\mathbb{R}}\frac{1}{\lvert x\rvert}\, dx=2\int_0^{\infty}\frac{1}{x}\, dx$ convergent?
– user34632
Commented Sep 25, 2013 at 16:50
• No, it is not. $\int_1^M\frac{1}{x}=\ln M \rightarrow \infty$ as $M\rightarrow \infty$. Commented Sep 25, 2013 at 16:59
• But I thought - I need a convergent majorant -
– user34632
Commented Sep 25, 2013 at 17:00
• And the minorant converges here??
– user34632
Commented Sep 25, 2013 at 17:05
• $\int_1^\infty \frac{x}{\sqrt{1+x^4}}$ dx is not convergent. You can't find a majorant convergent. Commented Sep 25, 2013 at 17:31

$$\frac{x}{\sqrt{1+x^4}} > \frac{x}{\sqrt{x^4+x^4}} > \frac{x}{2\sqrt{x^4}} = \frac{1}{2x} \text{ for }|x|>1, \text{ and } \int_{|x|>1} \frac{dx}{2x}=\infty.$$

• Sorry, my comment was just... weird. I'll remove it. I guess I forgot about that $x$ on top. Shit happens. Commented Sep 26, 2013 at 10:34