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Show that $$\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{x^{\frac{1}{2}}\ln x}{1+x^2}dx=\pi$$

After a number of transformations I ended having $$4\int_{0}^{\frac{1}{n^2}}\frac{\ln(x)}{x^4+1}dx$$ From here on I have no idea on how to continue.

Zacky
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  • That's wrong. $I=\frac{\pi^2}{2\sqrt{2}}$. –  Mar 03 '19 at 18:52
  • The integral $\displaystyle \int_0^\infty \frac{\sqrt{x}\ln x}{1+x^2},dx$ converges therefore your limit is the value of this integral. (whatever is $n$: integer, real) – FDP Mar 03 '19 at 19:56

2 Answers2

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Unfortunately your result is not the right one. Let us denote as follows:$$I=\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{x^{\frac{1}{2}}\ln x}{1+x^2}dx=\int_0^\infty \frac{\sqrt x \ln x}{1+x^2}dx\overset{x=t^2}=4\int_0^\infty \frac{t^2 \ln t}{1+t^4}dt\overset{t\rightarrow \frac{1}{y}}=-4\int_0^\infty \frac{\ln y}{1+y^4}dy$$ Now it is enough to use this general integral from here.

$$\int_0^\infty\frac{\log(x)}{1+x^n}dx=-\frac{\pi^2}{n^2}\csc\left(\frac\pi n\right)\cot\left(\frac\pi n\right)\Rightarrow \boxed{I=4 \cdot\frac{\pi^2}{4^2}\csc\left(\frac\pi 4\right)\cot\left(\frac\pi 4\right)=\frac{\pi^2}{2 \sqrt 2}}$$

Community
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Zacky
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    Another general integral for $a>0$: $\int_0^{\infty} \frac{\sqrt{x}\log(x)}{x^2+a^2} dx = \frac{\pi}{\sqrt{2a}}\left(\log(a)+\frac{\pi}{2}\right)$. Put in $a=1$. :) –  Mar 03 '19 at 19:14
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Here is an approach that will make use of the derivative of the Beta function $\operatorname{B} (m,n)$ of the form $$\operatorname{B} (m,n) = \int_0^\infty \frac{x^{m - 1}}{(1 + x)^{m + n}} \, dx, \quad m,n > 0.$$

Let $$I = \int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx.$$ Enforcing a substitution of $x \mapsto \sqrt{x}$ leads to $$I = \frac{1}{4} \int_0^\infty \frac{x^{-1/4} \ln x}{1 + x} \, dx.$$

Now consider $$J(s) = \frac{1}{4} \int_0^\infty \frac{x^s}{1 + x} \, dx, \qquad - 1 < s < 0.$$ On differentiating with respect to $s$ we have $$J'(s) = \frac{1}{4} \int_0^\infty \frac{x^s \ln x}{1 + x} \, dx.$$ For our integral we see that $I = J'(-1/4)$. Now \begin{align} J(s) &= \frac{1}{4} \int_0^\infty \frac{x^{(s + 1) - 1}}{(1 + x)^{(s + 1) + (-s)}} \, dx\\ &= \frac{1}{4} \operatorname{B} (s + 1, -s)\\ &= \frac{1}{4} \Gamma (s + 1) \Gamma (-s). \end{align} From the reflexion formula for the Gamma function, namely $$\Gamma (1 - x) \Gamma (x) = \frac{\pi}{\sin (\pi x)},$$ replacing $x$ with $-s$ gives $$\Gamma (s + 1) \Gamma (-s) = - \pi \operatorname{cosec} (\pi s).$$ Thus $$J(s) = -\frac{\pi}{4} \operatorname{cosec} (\pi s).$$ Differentiating with respect to $s$ gives $$J'(s) = \frac{\pi^2}{4} \cot (\pi s) \operatorname{cosec} (\pi s).$$ Setting $s = -1/4$ gives $$J \left (-\frac{1}{4} \right ) = \frac{\pi^2}{4} \cot \left (-\frac{\pi}{4} \right ) \operatorname{cosec} \left (-\frac{\pi}{4} \right ),$$ or $$\int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx = \frac{\pi^2}{2 \sqrt{2}}.$$

omegadot
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