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I found this thing in the "Table of Integrals, Series and Products" p. 569 $$I=\int_{-\infty}^{+\infty} \frac{\ln (\sin^2 x)}{1+x^2} dx = 2 \pi \ln \left( \frac{1-e^{-2}}{2}\right)$$

My first attempt.

I noticed that the integrand has discontinuities at pi-fold points, so I assumed that it makes sense to try the Lobachevsky formula or something like it. I tried to take an idea from it and this is what I got:

$$ I=\sum_{n\ \in\ \mathbb{Z}} \int_{\pi n}^{\pi n + \pi} \frac{\ln (\sin^2 x)}{1+x^2} dx = |x=t+\pi n|= \sum_{n\ \in\ \mathbb{Z}} \int_{0}^{\pi} \frac{\ln (\sin^2 t)}{1+(t+\pi n)^2} dt = $$ $$ =\int_{0}^{\pi} dt \ln (\sin^2 t) \sum_{n\ \in\ \mathbb{Z}} \frac{1}{1+(t+\pi n)^2} $$

The sum of the series is calculated using the properties of complex numbers and the digamma function.

$$ \sum_{n\ \in\ \mathbb{Z}} \frac{1}{1+(t+\pi n)^2}= \frac{\left(\cos2t+\cosh2\right)\sinh2}{\sin^{2}2t+\sinh^{2}2} $$

Replacing the series with this monster we get a horrible looking integral:

$$ I= \int_{0}^{\pi} \ln (\sin^2 t) \frac{\left(\cos2t+\cosh2\right)\sinh2}{\sin^{2}2t+\sinh^{2}2} dt $$

I have no idea what i should do with it.

My second attempt.

After a few unsuccessful substitutions, I assumed that a complex analysis can help me. I constructed a contour with notches at all special points except i. But I've discovered one unpleasant thing: Domain coloring of the function

The problem is that the function goes to another sheet, which makes it impossible to just take a half-circle contour. I tried combining both ways: I split the integral into a sum and tried calculating the pieces separately using residues, but it's worse there.

How to calculate it?

Rezerd Prime
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1 Answers1

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Note that the integrand is an even function. Therefore:

$$ 2 \int_{0}^{+\infty} \frac{\ln(\sin^2 x)}{1 + x^2} \, dx. $$

Recall the Fourier series expansion for $\ln(\sin x)$ on the interval $(0, \pi)$:

$$ \ln(\sin x) = -\ln 2 - \sum_{k=1}^{\infty} \frac{\cos(2k x)}{k}. $$

Hence:

$$ \ln(\sin^2 x) = -2\ln 2 - 2 \sum_{k=1}^{\infty} \frac{\cos(2k x)}{k}. $$

Substituting the expansion into the integral:

$$ 2 \int_{0}^{+\infty} \frac{-2\ln 2 - 2 \sum_{k=1}^{\infty} \frac{\cos(2k x)}{k}}{1 + x^2} \, dx, $$

which simplifies to:

$$ -4 \ln 2 \int_{0}^{+\infty} \frac{dx}{1 + x^2} - 4 \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{+\infty} \frac{\cos(2k x)}{1 + x^2} \, dx. $$

The first integral is:

$$ \int_{0}^{+\infty} \frac{dx}{1 + x^2} = \frac{\pi}{2}. $$

Using the standard integral:

$$ \int_{0}^{+\infty} \frac{\cos(a x)}{1 + x^2} \, dx = \frac{\pi}{2} e^{-a}, $$

for $a > 0$, and then, for $a = 2k$:

$$ \int_{0}^{+\infty} \frac{\cos(2k x)}{1 + x^2} \, dx = \frac{\pi}{2} e^{-2k}. $$

Substituting these results back:

$$ -4 \ln 2 \cdot \frac{\pi}{2} - 4 \sum_{k=1}^{\infty} \frac{1}{k} \cdot \frac{\pi}{2} e^{-2k} = -2\pi \ln 2 - 2\pi \sum_{k=1}^{\infty} \frac{e^{-2k}}{k}. $$

Recognizing the series as a logarithmic function:

$$ \sum_{k=1}^{\infty} \frac{e^{-2k}}{k} = -\ln(1 - e^{-2}), $$

the integral becomes:

$$ -2\pi \ln 2 + 2\pi \ln(1 - e^{-2}) = 2\pi \ln \left(\frac{1 - e^{-2}}{2} \right). $$

PaulS
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