Stuck and confused with $\int \frac{u^{1/2}}{(u+1)^2} du$

Question

I'm trying to find the expectation value $\langle x^2 \rangle_{\psi}$ for

$$\psi(x,0) = \sqrt{\frac{2}{\pi}}a \cdot \frac{1}{x^2+a^2}$$

and I know the result is $a^2$. I'm terribly stuck integrating this. Currently:

$$\langle x^2 \rangle_{\psi} = \int\limits_{-\infty}^{\infty}\, \psi^* x^2 \psi \, dx$$ $$= \left(\frac{2a^2}{\pi}\right) \int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^2} dx = \left(\frac{2a^2}{\pi}\right) \int_{-\infty}^{\infty} \frac{1}{a^2} \frac{(x/a)^2}{((x/a)^2+1)^2}dx = \left(\frac{4a}{\pi}\right) \int^\infty_0 \frac{u^2 }{(u^2+1)^2} du$$ where the change of variable $u=x/a$ is used.

With $v=u^2$, $u=\sqrt{v}$ and $du=\tfrac{dv}{2\sqrt{v}}$ and then, $$\langle x^2 \rangle_{\psi} = \left(\frac{2a}{\pi}\right) \int^\infty_0 \frac{v^{1/2}}{(v+1)^2}dv = \frac{2a}{\pi} \int^\infty_0 \frac{v}{(v+1)^2}v^{1/2} \frac{dv}{v}$$

Then I checked the latter integral in Wolfram Mathematica but the results doesn't seem to match (not $\pi a$) and this is confusing me a lot. Any suggestions with this?

Also: is it possible to solve it analytically?

Answer 1

The term $\langle x^2\rangle_\psi$ as defined by the OP can be obtained through standard method of integration of rational functions.

Here is one method:

$$ \int \frac{x^2}{(x^2+a^2)^2}\,dx =\int \frac{dx}{x^2+a^2} -a^2\int\frac{dx}{(x^2+a^2)^2} $$

Setting $I_n=\int\frac{dx}{(x^2+a^2)^n}$, we get by integration by parts the recurrence formula $$ I_{n+1}=\frac{1}{2na^2}\frac{x}{(x^2+a^2 )^n}+\frac{2n-1}{2na^2}I_n$$ For $n=1$ we obtain $$ I_2=\frac{1}{2a^2}\frac{x}{x^2+a^2} +\frac{1}{2a^3}\arctan(x/a) + C $$ Putting things together, we het

$$ \frac{2a^2}{\pi}\int^\infty_{-\infty}\frac{x^2\,dx}{(x^2+a^2)^2}=\frac{2a^2}{\pi}\Big(\frac{1}{2a}\arctan(x/a)-\frac{1}{2}\frac{x}{x^2+a^2}\Big)\Big|^\infty_{-\infty}=a $$

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The OP, through a couple of change of variables, reduced the quantity of interest to calculating the integral $$\int^\infty_0 \frac{u^{1/2}}{(u+1)^2}\,du=\int^\infty_0 \frac{u}{(u+1)^2}u^{1/2}\,\frac{du}{u}$$ This integral can be obtained by calculus of residues. In this instance, the Mellin transform comes to mind (See Marsden's basic complex analysis, or Lang's complex analysis).

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The Mellin transform of a function $f$ is defines as $M_f(s)=\int^\infty_0 f(x) x^s\,\frac{dx}{x}$, $s>0$. Under the following conditions

1. $f$ is analytic on the complex plane, except for a finite number of poles none of which lies on the positive real axis.
2. There are $0<c<s<b$ such that $f(z)=O(z^{-b})$ as $z\rightarrow\infty$, and $f(z)=O(z^{-c})$ as $z\rightarrow0$ we have that $$\int^\infty_0 f(x) x^{s-1}\,dx = -\frac{\pi e^{-\pi i s}}{\sin \pi s}\sum^N_{j=1} \operatorname{Res}_{z_j}(f(z) z^{s-1})$$ where $z_1,\ldots, z_N$ are the poles of $f$ in $\mathbb{C}\setminus\{0\}$, and $z^{(s-1)}=e^{(s-1)\log z}$ with $\log$ is the logarithmic function with branch $0<\operatorname{arg}(z <2\pi$.

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For the integral in the OP, it is easy to check that $f(z)=\frac{z}{(z+1)^2}$ satisfies condition (1) and (2) with $s=1/2$.

$g(z)=f(z) z^{1/2 -1}$ has a pole at $z=-1$, and

$$\operatorname{Res}_{-1}(g)=1\cdot e^{-\tfrac12\pi i} -(\tfrac12-1)e^{-\tfrac32 \pi i}=-\tfrac{i}{2}$$

Hence $$ \int^\infty_0 \frac{u^{1/2}}{(u+1)^2}\,du =-\frac{\pi e^{-\tfrac12\pi i}}{\sin\tfrac12 \pi}(-\tfrac{i}{2})=\frac{\pi}{2} $$ Thus the quantity $\langle x^2\rangle_{\psi}$, as defined by the OP, is $\frac{2a}{\pi}\frac{\pi}{2}=a$.

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Edit: It seems that the OP is missing a factor $a$ in the normalization factor. $$ \begin{align} 1=\int^\infty_{-\infty}\psi^*\psi&=c\int^\infty_{-\infty}\frac{dx}{(x^2+a^2)^2}\\ &=c\Big( \frac{1}{2a^2}\frac{x}{(x^2+a^2)^2}+\frac{1}{2a^3}\arctan(x/a)\Big)\Big|^\infty_{-\infty}\\ &=\frac{\pi}{2a^3}c \end{align} $$ implies that $c=\frac{2a^3}{\pi}$. The OP has $\frac{2a^2}{\pi}$

Answer 2

Hint:

By parts,

$$\int\frac{u^2}{(1+u^2)^2}du=\int u\frac u{(1+u^2)^2}du=-u\frac 1{2(1+u^2)}+\frac12\int\frac{du}{1+u^2} \\=-\frac u{2(1+u^2)}+\frac12\arctan(u).$$

Answer 3

If you want to solve it with a substitution, instead of $v=u^2$ take $u=\tan t$, so$$\int_0^\infty\frac{u^2du}{(u^2+1)^2}=\int_0^{\pi/2}\sin^2tdt=\frac{\pi}{4}.$$