I am trying to evaluate the following integral:
$$\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx,$$ where $a>b>0$.
I can't really think of a way to find it. So, please give me a hint.
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you could do the following: write $\log((ax)^2+bx+a)$ as $\log([x-x_-])+\log([x-x_+])$ and use partial fraction for $\frac{1}{(1+x^2)}$. Afterwards integrate every of the four resulting terms separately. Unfortunately the result will look messy anyways :(. – tired Jan 21 '15 at 14:26
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With $\sin\theta =\frac ba$\begin{align} &\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx\\ =& \int^{1}_{-1} \frac{\ln a}{x^2+1}dx+ \int^{1}_{-1} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}\overset{x\to1/x}{dx}\\ =& \ \frac\pi2\ln a+\frac12 \int^{\infty}_{-\infty} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}dx-2\int_1^\infty \frac{\ln x}{x^2+1}dx\\ = &\ \frac\pi2\ln a+\frac\pi2 \ln\left(2\cos\frac\theta2\right)-2G\\ =& \ \frac\pi2\ln2 +\frac\pi2 \ln\left( \sqrt{a^2-b^2}+a\right)-2G \end{align}
Quanto
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how did you made the limits to infinity? sorry if i am missing some theorems – hemant kumar Jan 08 '23 at 17:30
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1@hemantkumar - first $\int_{-1}^1= \int_{-1}^0 + \int_{0}^1$, then $x\to 1/x$ for both – Quanto Jan 08 '23 at 17:33
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Why $\sin (\theta) = b/a$ ? How you know $b/a$ belongs to $[-1,1]$? – An_Elephant Jan 08 '23 at 17:54
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@lonestudent Suppose $a=2$ and $b=1$ . Then $b/a$ does not equal $\sin \theta$? – An_Elephant Jan 08 '23 at 17:58
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@Quanto how did you integrated $\frac{1}{2}\int_{-\infty}^{+\infty}$ to get $\frac{\pi}{2}ln(2cos(\theta/2))$ – hemant kumar Jan 08 '23 at 18:09
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1@hemantkumar - Feymann’s trick: $I(\theta)= \int^{\infty}_{-\infty} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}dx$ and $I’(\theta)=-\pi\tan\frac\theta2$ – Quanto Jan 08 '23 at 18:21
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