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Use the substitution $u = \frac{1}{x}$ to show that $\int_{0}^{\infty} \frac{\ln(x)}{1+x^2} dx = 0$

I know that this is improper and I could rewrite this as

$$\lim_{A\to0^+} \int_{A}^{1} f(x)dx + \lim_{B\to\infty} \int_{1}^{B} f(x)dx$$

Now I'm currently trying to figure out the integration of this

Using u = $\frac{1}{x}$, then $-du = \frac{1}{x^2} dx$. We also have $x = \frac{1}{u}$

$$= - \int\frac{ln \left(\frac{1}{u}\right)}{1+ \frac{1}{u^2}}\frac{1}{u^2}du = -\int \frac{\ln(\frac{1}{u})}{u^2+1}du$$

No idea how to do do this after that.

Shaun
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user349557
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    After your last line, use $$-\ln(1/u)=\ln u$$ (For your future use, please prefer \ln over ln.) – Did Apr 16 '17 at 08:30

1 Answers1

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Note that we have convergence of both improper integrals

$$\int_1^\infty \frac{\ln x}{1+x^2} \, dx, \,\,\, \int_0^1 \frac{\ln x}{1+x^2} \, dx.$$

The first can be shown to converge by the comparison test using

$$\frac{\ln x}{1+x^2} = \frac{2\ln \sqrt{x}}{1+x^2} \leqslant \frac{2\ln \sqrt{x}}{x^2} \leqslant \frac{2\sqrt{x}}{x^2} = \frac{2}{x^{3/2}}.$$

The second convergence follows from $\ln x /(1 + x^2) \sim \ln x$ as $x \to 0$ and

$$\int_0^1 \ln x \, dx = \left.(x \ln x - x)\right|_0^1 = -1.$$

After establishing the existence of the integrals, the trick is to show

$$\int_0^1 \frac{\ln x}{1 + x^2} \, dx = -1 \cdot\int_1^\infty \frac{\ln x}{1 + x^2} \, dx, $$

With $u = 1/x$ we have

$$\int_1^\infty \frac{\ln x}{1 + x^2} \, dx = \int_1 ^0 \frac{\ln u^{-1}}{1 + u^{-2} }(-u^{-2}) du \\= -\int_1^0 \frac{-\ln u}{1 + u^2} \, du \\ = -\int_0^1 \frac{\ln u}{1 + u^2} \, du $$

RRL
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