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.

  • 2
    $\begingroup$ 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

Let $u=1/y^2$ and


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


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

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


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

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 $f'(u)$.


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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