st3he1d0t

2022-09-01

How can a positive integrand integrate to 0?

I integrated $\frac{\mathrm{log}x}{1+{x}^{2}}$ from $0$ to infinity with residue calculus and got... $0$

This also agrees with Wolfram Alpha.

How can this be?

Is it due to the behavior of $\mathrm{log}(x)$ near the origin? Like a cancellation effect?

Thanks,

I integrated $\frac{\mathrm{log}x}{1+{x}^{2}}$ from $0$ to infinity with residue calculus and got... $0$

This also agrees with Wolfram Alpha.

How can this be?

Is it due to the behavior of $\mathrm{log}(x)$ near the origin? Like a cancellation effect?

Thanks,

Matthew Benton

Beginner2022-09-02Added 9 answers

After the substitution $u=\frac{1}{x}$ you get

$${\int}_{1}^{\mathrm{\infty}}\frac{\mathrm{log}x}{1+{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{1}^{\mathrm{\infty}}\frac{\mathrm{log}(\frac{1}{x})}{1+\frac{1}{{x}^{2}}}\cdot (-\frac{1}{{x}^{2}})\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{1}^{0}\frac{\mathrm{log}u}{1+{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du=-{\int}_{0}^{1}\frac{\mathrm{log}u}{1+{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du$$

Yes, the two integrals cancel out.

If you calculate your integral the same way, you get that it is equal to minus itself...

$${\int}_{1}^{\mathrm{\infty}}\frac{\mathrm{log}x}{1+{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{1}^{\mathrm{\infty}}\frac{\mathrm{log}(\frac{1}{x})}{1+\frac{1}{{x}^{2}}}\cdot (-\frac{1}{{x}^{2}})\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{1}^{0}\frac{\mathrm{log}u}{1+{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du=-{\int}_{0}^{1}\frac{\mathrm{log}u}{1+{u}^{2}}\phantom{\rule{thinmathspace}{0ex}}du$$

Yes, the two integrals cancel out.

If you calculate your integral the same way, you get that it is equal to minus itself...

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