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$$\int_1^{\infty} \frac1x \left( \int_{x}^{2x} \frac{1}{1+t^2}dt \right) dx \leqslant \int_1^{\infty} \frac1x \left( \int_{x}^{2x} \frac{1}{1+x^2}dt \right) dx$$ and $$\int_1^{\infty} \frac1x \left( \int_{x}^{2x} \frac{1}{1+x^2}dt \right) dx = \int_1^{\infty} \frac1{1+x^2} dx $$ I'm trying to understand these two equations. Could somebody explain why do they stand.

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up vote 1 down vote accepted

If $x\geq 1$ is fixed, then $$ 0\leq \frac{1}{1+t^2}\leq \frac{1}{1+x^2},\quad \text{for }\; x\leq t\leq 2x, $$ and hence $$ \int_x^{2x} \frac{1}{1+t^2}\,\mathrm dt \leq \int_x^{2x}\frac{1}{1+x^2}\,\mathrm dt. $$ Now, multiply both sides by $\frac1x$ and integrate over $[1,\infty)$ with respect to $x$.

For the last equality, note that $$ \int_x^{2x}\frac{1}{1+x^2}\,\mathrm dt=\frac{1}{1+x^2}\int_x^{2x}1\,\mathrm dt=\frac{1}{1+x^2}(2x-x)=\frac{x}{1+x^2}. $$ Multiply this by $\frac1x$ and integrate over $[1,\infty)$ with respect to $x$.

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