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I have been trying to solve this integral for some time now

$\int\limits ^{0}_{-\infty }\frac{\ln( t+1)}{t^{2} +1} dt$.

and all the calculators I've used say it's equal to $\frac\pi4\ln(2)-G+\frac{π^{2}i}{4}$ ($G$ is Catalan's Constant), but I find that it's equal to $\frac\pi4\ln(2)-G+\frac{3π^{2}i}{4}$.

I can't seem to find the error, if any, in what I've done $\int\limits ^{0}_{-\infty }\frac{ln( t+1)}{t^{2} +1} dt=\int\limits ^{\infty }_{0}\frac{ln( -x+1)}{x^{2} +1} dx=\int\limits ^{\infty }_{0}\frac{ln( x-1)}{x^{2} +1} dt+\frac{π^{2} i}{2} =-G+\frac{π}{4} ln( 2) +\frac{3π^{2} i}{4}$.

Any help would be greatly appreciated.

Spador Yedi
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1 Answers1

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Your error is in applying the identity $$\ln{(-x)}=\ln{(x)}+i\pi\quad\forall x\in(0,\infty)$$ when $x\lt0$. You should get $$\begin{align} \int_0^\infty\frac{\ln{(-x+1)}}{x^2+1}\mathrm{d}x &=\int_0^1\frac{\ln{(-x+1)}}{x^2+1}\mathrm{d}x+\int_1^\infty\frac{\ln{(-x+1)}}{x^2+1}\mathrm{d}x\\ &=\int_0^1\frac{\ln{(x-1)}-i\pi}{x^2+1}\mathrm{d}x+\int_1^\infty\frac{\ln{(x-1)}+i\pi}{x^2+1}\mathrm{d}x\\ &=\int_0^\infty\frac{\ln{(x-1)}}{x^2+1}\mathrm{d}x+\left[-i\pi\arctan{(x)}\right]_0^1+\left[i\pi\arctan{(x)}\right]_1^\infty\\ &=\int_0^\infty\frac{\ln{(x-1)}}{x^2+1}\mathrm{d}x\\ \end{align}$$

Peter Foreman
  • 19,947