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In the book Sequential Analysis of Siegmund there is a calculation of an expectation on page 225. I am stuck at the very beginning when he notes that by the change of variable $\lambda=\alpha e^{u}$ that $$ \log\alpha=(2\pi)^{-1}\int_{-\infty}^{\infty}\frac{\alpha}{\alpha^{2}+\lambda^{2}}\log(\lambda^{2})d\lambda $$

I put this integral in the integral calculator and I get something rather unpleasant and I do not know what to do at this point. I also tried changing the variable as indicated but it was a dead end for me.

Thanks for the help

Gary
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Omer
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  • Have you tried a residue theorem approach? – Gary Jul 03 '23 at 07:25
  • @Gary I have not indeed. The residue is at infinity right? So I have to calculate the $\lim_{\lambda\rightarrow\infty}\lambda\cdot f(\lambda)$ if $f(\lambda)$ denotes the integrand. – Omer Jul 03 '23 at 07:57
  • The integral can be written as $\frac{2\alpha}{\pi}\int_{0}^{\infty}\frac{\ln x}{\alpha^{2}+x^{2}}dx$. There are some duplicates of this, like here. – Accelerator Jul 03 '23 at 09:47

2 Answers2

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Consider instead

$$I = \int_0^\infty \frac{2\alpha \log \lambda}{\alpha^2+\lambda^2}d\lambda$$

Use the substitution $\lambda = \frac{\alpha^2}{t}$:

$$I = \int_0^\infty \frac{2\alpha^3(\log \alpha^2 - \log t)}{\alpha^2t^2+\alpha^4}dt = \int_0^\infty \frac{4\alpha\log \alpha}{\alpha^2+t^2}dt - I$$

Solving for $I$ gives us

$$I = \int_0^\infty \frac{2\alpha\log \alpha}{\alpha^2+t^2}dt = \pi\log\alpha$$

The integral we want is $\frac{1}{2\pi}\cdot 2 \cdot I = \frac{1}{\pi}I$ by even function symmetry, therefore

$$\frac{1}{2\pi}\int_{-\infty}^\infty \frac{\alpha \log \lambda^2}{\alpha^2+\lambda^2}d\lambda = \log\alpha$$

Omer
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Ninad Munshi
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Letting $\lambda=\alpha \tan \theta$, then $$ I=2 \int_0^{\frac{\pi}{2}} \ln \left(\alpha^2 \tan ^2 \theta\right) d \theta= 2\left(\int_0^{\frac{\pi}{2}} 2 \ln \alpha d\theta+\int_0^{\frac{\pi}{2}} 2 \ln (\tan \theta) d \theta\right)=2\pi\ln \alpha $$

Lai
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