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$\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$

Now Here $$\displaystyle I_{2} = \int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx.$$ Put $\displaystyle x=\frac{1}{t}.$ Then $\displaystyle dx = -\frac{1}{t^2}dt$ and changing limit

We get $$\displaystyle I_{2} = \int_{1}^{0}\frac{-t^3\cdot \ln t}{(1+t^2)^2}\cdot -\frac{1}{t^2}\,dt = \int_{1}^{0}\frac{t\ln t}{(1+t^2)^2}\,dt = -\int_{0}^{1}\frac{t\ln t}{(1+t^2)^2}\,dt$$

So we get $$\displaystyle I_{2} = -\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}dt = -I_{1}\Rightarrow I_{1}+I_{2} = 0$$

So we get $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx = I_{1}+I_{2} =0$$

My question is, can we solve it using another method? If so then please explain.

gebruiker
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juantheron
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4 Answers4

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Your solution is simple and straight to the point. Here is a longer way:

You could take the indefinite integral then apply limits to find the definite integral.

Let $v=\log x$, $u'=\frac{x}{(1+x^2)^2}$

So $v'=\frac{1}{x}$, $u=-\frac{1}{2(1+x^2)}$ $$\int\frac{{x\log x}}{(1+x^2)^2}dx=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x(1+x^2)}dx$$ $$=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x}-\frac{x}{2(1+x^2)}dx$$ $$=-\frac{\log x}{2(1+x^2)}+\frac{\log x}{2}-\frac{\log(1+x^2)}{4}$$ $$=-\frac{x^2\log x}{2(1+x^2)}-\frac{\log(1+x^2)}{4}$$

Ian Miller
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$$\int\frac{x\ln(x)}{(x^2+1)^2}\space\text{d}x=$$

Integrate by parts $\int f\text{d}g=fg-\int g\text{d}f$: $f=\ln(x),\text{d}g=\frac{x}{(x^2+1)^2}\space\text{d}x, \text{d}f=\frac{1}{x},g=\frac{1}{2(x^2+1)}$:

$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\frac{1}{x(x^2+1)}\space\text{d}x=$$

Substitute $u=x^2$ and $\text{d}u=2x\space\text{d}x$:

$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\int\frac{1}{u(u+1)}\space\text{d}u=$$ $$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\int\left(\frac{1}{u}-\frac{1}{u+1}\right)\space\text{d}u=$$ $$-\frac{\ln(x)}{2(x^2+1)}-\frac{1}{4}\int\left(\frac{1}{u+1}-\frac{1}{u+1}\right)\space\text{d}u+\frac{1}{4}\int\frac{1}{u}\space\text{d}u=$$

Substitute $s=u+1$ and $\text{d}s=\text{d}u$:

$$-\frac{\ln(x)}{2(x^2+1)}-\frac{1}{4}\int\left(\frac{1}{u+1}-\frac{1}{s}\right)\space\text{d}s+\frac{1}{4}\int\frac{1}{u}\space\text{d}u=$$ $$-\frac{\ln(x)}{2(x^2+1)}-\frac{\ln(s)}{4}+\frac{\ln(u)}{4}+\text{C}=$$ $$-\frac{\ln(x)}{2(x^2+1)}-\frac{\ln(x^2+1)}{4}+\frac{\ln(x^2)}{4}+\text{C}=$$ $$\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)+\text{C}$$

Setting the boundaries:

  • Zero: $$\lim_{x\to 0}\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=$$ $$\frac{1}{4}\lim_{x\to 0}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=$$ $$\frac{1}{4}\left(2\lim_{x\to 0}x^2\ln(x)-\lim_{x\to 0}\ln(x^2+1)\right)=$$ $$\frac{1}{4}\left(2\cdot 0-0\right)=0$$
  • Infinity: $$\lim_{x\to\infty}\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)=0$$

So:

$$\int_{0}^{\infty}\frac{x\ln(x)}{(x^2+1)^2}\space\text{d}x=\left[\frac{1}{4}\left(\frac{2x^2\ln(x)}{x^2+1}-\ln(x^2+1)\right)\right]_{0}^{\infty}=0-0=0$$

Jan Eerland
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1

The simplest solution for this problem is to use the substitution $x=\frac1t$. In fact \begin{eqnarray} I&=&\int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx I&=&\int_{\infty}^0\frac{\frac{1}{t}\ln \frac{1}{t}}{(1+(\frac{1}{t^2})^2}(-\frac{1}{t^2})dt\\ &=&-\int_{0}^{\infty}\frac{t\ln t}{(1+t^2)^2}dt\\ &=&-I \end{eqnarray} and hence $I=0$.

xpaul
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1

$$I= \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx=\int_{0}^{\infty}\frac{x\ln x}{x^4+2x^2+1}dx$$

$${\color{blue}{F(s,\alpha)=\int_0^\infty \frac{x^{s-1}}{x^2+2x\cos\alpha +1}\,dx=\frac{\pi\sin((1-s)\alpha)}{\sin\alpha\sin(\pi s)}}}$$

Proof of above formula

$${\underset{x\to x^2}\implies{F(s,\alpha)=\int_0^\infty \frac{x^{2s-1}}{x^4+2x^2\cos\alpha +1}\,dx=\frac{\pi\sin((1-s)\alpha)}{2\sin\alpha\sin(\pi s)}}}$$

$$ \boxed{\frac{\partial}{\partial s}F\left(s,\alpha\right)= \int_0^\infty \frac{x^{2s - 1} \ln x}{x^4 + 2x^2\cos\alpha + 1} \, dx = \frac{\pi}{4 \sin\alpha} \frac{ -\alpha \cos((1-s)\alpha) \sin(\pi s) - \pi \sin((1-s)\alpha) \cos(\pi s)}{ \sin^2(\pi s) } }$$

$${\therefore{I=\lim_{\alpha\to 0}\lim_{s\to 1}}\frac{\pi}{4 \sin\alpha} \frac{ -\alpha \cos((1-s)\alpha) \sin(\pi s) - \pi \sin((1-s)\alpha) \cos(\pi s)}{ \sin^2(\pi s) }=0}$$

Amrut Ayan
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