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I'm trying to calculate such a limit:

$$ \lim_{y \to \infty} \int_{\mathbb{R}} \frac{xy^2}{(y-x)^2+y^2}\ln{\left(1+\frac{1}{x^2}\right)}\mbox{d}x$$

My main idea was to make a substitution $x = ys$, so the limit looks like:

$$\lim_{y \to \infty} \int_{\mathbb{R}} \frac{sy^2}{(1-s)^2+1}\ln{\left(1+\frac{1}{(ys)^2}\right)}\mbox{d}s$$.

Now, if I look at the first term of Taylor expansion for logarithm, I get just

$$\int_{\mathbb{R}} \frac{1}{s((1-s)^2+1)}\mbox{d}s$$

which happens to have finite principal value (at least...). But the other terms are quite problematic near $s=0$ and I cannot handle that, so I think Taylor expansion is not a good idea here. I also tried some integration by parts, but it didn't work.

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    $\begingroup$ Mathematica can solve the $x$-integral from $-\infty$ to $+\infty$ and gives $\frac{\pi}{2}$ for the limit. $\endgroup$ – Dr. Wolfgang Hintze Aug 4 '20 at 9:55
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Let $u=1/y^2$ and

$$f(u)=\int_{-\infty}^\infty{s\over(s-1)^2+1}\ln(1+u/s^2)\,ds$$

Note first that $s\ln(1+u/s^2)\to0$ both as $s\to0$ and as $s\to\pm\infty$, so the improper integral converges for all $u\ge0$, and, by dominated convergence, we have $\lim_{u\to0^+}f(u)=f(0)=0$. The limit we need to evaluate is $\lim_{u\to0^+}{f(u)\over u}$. L'Hopital tells us this is equal to $\lim_{u\to0^+}f'(u)$, provided that limit exists.

Working formally at first, we have

$$f'(u)=\int_{-\infty}^\infty{s\over(s-1)^2+1}\cdot{1\over s^2+u}\,ds$$

which also converges as long as $u$ is positive. (Note: If we let $u=0$ in this formula for $f'(u)$, the integrand has a pole at $s=0$ and the improper integral does not converge, unless one takes extra care to give it a "principal value" interpretation. But L'Hopital doesn't care about value of the derivative at $0$, just the values near $0$.)

Partial fractions lets us compute the indefinite integral:

$${s\over((s-1)^2+1)(s^2+u)}={1\over u^2+4}\left({(u-2)(s-1)+u+2\over(s-1)^2+1}-{(u-2)s+2u\over s^2+u}\right)$$

so that

$$\begin{align} f'(s) &={u-2\over u^2+4}\int_{-\infty}^\infty\left({s-1\over(s-1)^2+1}-{s\over s^2+u} \right)\,ds+{1\over u^2+4}\int_{-\infty}^\infty\left({u+2\over(s-1)^2+1}-{2u\over s^2+u} \right)\,ds\\\\ &={u-2\over u^2+4}\cdot{1\over2}\ln\left((s-1)^2+1\over s^2+u \right)\Big|_{-\infty}^\infty+{(u+2)\arctan(s-1)-2\sqrt u\arctan s\over u^2+4}\,\Big|_{-\infty}^\infty\\\\ &={(u+2-2\sqrt u)\pi\over u^2+4} \end{align}$$

(in particular, the log term vanishes at $s=\pm\infty$), from which we see that

$$\lim_{u\to0^+}f'(u)={(0+2-2\sqrt0)\pi\over0^2+4}={\pi\over2}$$

and we are thus done, provided we justify the formalism of differentiating inside the integral. But this also comes courtesy of dominated convergence, since for any fixed positive value of $u$ and any appropriate small value of $h$ (so that $u+h$ is still positive), we have

$${f(u+h)-f(u)\over h}={1\over h}\int_{-\infty}^\infty{s\over(s-1)^2+1}\ln\left(1+{h\over s^2+u} \right)\,ds$$

and

$${1\over h}\left|{s\over(s-1)^2+1}\ln\left(1+{h\over s^2+u} \right) \right|\le{s\over((s-1)^2+1)(s^2+u)}$$

which, for any $u\gt0$, is integrable over $\mathbb{R}$. This lets us take the limit as $h\to0$ inside the integral sign, obtaining the asserted integral expression for $f'(u)$.

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Let $f \colon (0,\infty) \to (0,\infty),$ $$ f(y) = \int \limits_\mathbb{R} \frac{x y^2}{(y-x)^2 + y^2} \, \ln \left(1 + \frac{1}{x^2}\right) \, \mathrm{d} x \stackrel{x = \frac{1}{t}}{=} \int \limits_\mathbb{R} \frac{\ln \left(1 + t^2\right)}{t^2 + \left(\frac{1}{y} - t\right)^2} \, \frac{\mathrm{d} t}{t} = g \left(1, \frac{1}{y}\right)\, .$$ Here, $g \colon [0,\infty) \times (0,\infty) \to [0,\infty)$ is defined by $$ g(a,b) = \int \limits_\mathbb{R} \frac{\ln \left(1 + a^2 t^2\right)}{t^2 + \left(b - t\right)^2} \, \frac{\mathrm{d} t}{t} \, .$$ For $a, b >0$ we have $$ \partial_1 g(a,b) = 2 a \int \limits_\mathbb{R} \frac{t}{\left[1 + a^2 t^2\right] \left[t^2 + (b-t)^2\right]} \, \mathrm{d} t = \frac{2 \pi a}{1+ (1 + a b)^2} \, .$$ The integral can be evaluated using the residue theorem and the usual semi-circle contour. Since $g(0,b) = 0$ holds for $b > 0$, we find $$ f(y) = g \left(1, \frac{1}{y}\right) = \int \limits_0^1 \partial_1 g \left(a, \frac{1}{y}\right) \mathrm{d} a = 2 \pi \int \limits_0^1 \frac{a}{1 + \left(1+\frac{a}{y}\right)^2} \, \mathrm{d} a \, , \, y > 0 \, . $$ Now we can use the dominated convergence theorem to obtain $$ \lim_{y \to \infty} f(y) = 2 \pi \int \limits_0^1 \frac{a}{2} \, \mathrm{d} a = \frac{\pi}{2} $$ in agreement with Dr. Wolfgang Hintze's Mathematica result. The Taylor series of the integrand in $\frac{1}{y}$ yields the more precise asymptotic expansion $$ f(y) \sim \frac{\pi}{2} \left[1 - \frac{2}{3y} + \frac{1}{4y^2} + \mathcal{O} \left(\frac{1}{y^4}\right)\right] \, , \, y \to \infty \, .$$

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