Integral $\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^b}$

Question

For Quantum Mechanics I need to normalize two wavefunctions and I haven't been able to figure out the integrals. The integrals involve a normalization constant, and 2 flexible parameters since these wavefunctions are then passed to the variation principle to solve the upper bound to the groundstate energy. The Quantum behind the math isn't important in what I am trying to solve

The following are the integrals I cannot figure out, find in an integral table, or use wolframalpha to determine:

Note: 'A' is the normalization constant that can be Complex; what I need to solve for. 'a'and 'b' are the flexible parameters

$$ 1=|A|^2\int_{-\infty}^{\infty} (a^2+x^2)^{-b}dx $$ $$ 1=4\pi |A|^2\int_{0}^{\infty} (a+r)^{-2b}r^2dr $$

So in both cases I need to solve for A (or A squared) in terms of the flexible parameters 'a'and 'b'.

Answer 1

For the first normalization constant $$ |A|^2\int_{-\infty}^{\infty} (a^2+x^2)^{-b}dx=1, $$ note the integrand is even so we can write $$ |A|^2\int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx=\frac{1}{2}. $$ We can define this integral as $$ I_1\equiv \int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx, \quad \Re(a^2)>0, \Re(b)>\frac{1}{2}. $$ Note the constraints placed on $a,b$, otherwise your integral isn't convergent. Make sure you understand this from a physical standpoint as well as this is quantum mechanics. This integral can be calculated by first considering $$ I_{1,2}(a)\equiv\int_0^\infty \frac{1}{x^2+a^2}dx, $$ now differentiate this once w.r.t a to obtain $$ I'_{1,2}(a)=\int_0^\infty \frac{\partial}{\partial a}\frac{1}{x^2+a^2}dx=-2a\int_0^\infty \frac{dx}{(x^2+a^2)^2}. $$ If you generalize to differentiating n times, you can see the pattern easily. We obtain $$ I^{(n)}_{1,2}(a)=(-1)^{b-1}\dfrac{(2b-3)!!}{2^{b}a^{2b-1}}\cdot\pi. $$ We can integrate this result to obtain $I_1$ which is given by $$ I_1=\int_0^\infty \frac{1}{(x^2+a^2)^{b}}dx=\frac{1}{2\Gamma(b)} \sqrt{\frac{\pi}{a^2}} a^{2-2b}\Gamma\big(b-\frac{1}{2}\big), \quad \Re(a^2)>0, \Re(b)>\frac{1}{2}. $$ Thus your first normalization integral gives us $$ |A|^2\int_{0}^{\infty} (a^2+x^2)^{-b}dx=\frac{1}{2}=|A|^2\frac{1}{2\Gamma(b)} \sqrt{\frac{\pi}{a^2}} a^{2-2b}\Gamma\big(b-\frac{1}{2}\big)=\frac{1}{2}, $$ you can solve for A. I am still working on your second normalization integral.

Your second normalization integral given by $$ 1=4\pi |A|^2\int_{-\infty}^{\infty} (a+r)^{-2b}r^2dr\equiv 4\pi |A|^2 I_2. $$ We now calculate $I_2$ which is given by $$ I_2=\int_{-\infty}^{\infty} \frac{r^2}{(a+r)^{2b}}dr=0. $$ However this integral evaluates to 0 because of the lower limit of integration. I think you meant to write $$ I_2=\int_{0}^{\infty} \frac{r^2}{(a+r)^{2b}}dr=\frac{a^{3-2b}}{4b^3-12b^2+11b-3}, \Re(a)>0, 2\Re(b)>3, \arg(a^{-1})\leq \pi. $$ The integral should be from $0$ to $\infty$ because the radius is always positive in physics. Thus your second normalization constant can be calculated by $$ 1=4\pi |A|^2\cdot \frac{a^{3-2b}}{4b^3-12b^2+11b-3}, $$ solve for A. Let me know what you think, and if you need more help. Also, clarify with me that you agree you posted the second normalization integral incorrectly because of the lower bound you wrote as $-\infty$.

UPDATE: Gamma Functions in elementary quantum mechanics $\Gamma(n)=(n-1)!$.

Also, in regards to Hydrogen $$ \int_0^\infty x^n e^{-x} dx=n!=\Gamma(n+1), \quad \Re(n) >-1 $$ this is what $\big<r^n\big>$ is equal to if we are discussing the ground state of Hydrogen $\psi \propto e^{-r/a}$. Common thing to do is use a Gaussian $\psi \propto e^{-r^2/a^2}$ and use the variational principle to estimate the Hydrogen ground state energy. You will see the difference. Another quantity that comes up often is $\big<x^{2p}\big>$ for the quantum harmonic oscillator. We need integrals of the form (ground state of oscillator) $$ \int_{-\infty}^\infty e^{-x^2}x^{2p}dx=\frac{1}{2}\big(1+(-1)^{2p}\big) \Gamma\big(\frac{1}{2}+p\big), \quad \Re(p) > -\frac{1}{2}. $$ Put p=1, and see what you get by hand (think about it physically) and check it with this formula!

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\cal I}\pars{\mu,\nu} \equiv \int_{0}^{\infty}\pars{1 + \mu x^{2}}^{-\nu}\,\dd x\,,\quad\mu > 0,\ \nu > {3 \over 4} \quad\mbox{and}\quad{\cal I}\pars{\mu,\nu} = \mu^{-1/2}\int_{0}^{\infty}\pars{1 + x^{2}}^{-\nu}\,\dd x.\ \mbox{Also}, \partiald{{\cal I}\pars{\mu,\nu}}{\mu}=-\nu\int_{0}^{\infty}\pars{1 + \mu x^{2}}^{-\nu - 1}x^{2}\,\dd x}$

With $\ds{t \equiv {1 \over 1 + x^{2}}\quad\imp\quad x = \pars{{1 \over t} - 1}^{1/2}}$: \begin{align} \color{#00f}{\large{\cal I}\pars{\mu,\nu}}&= \mu^{-1/2}\int_{0}^{1} t^{\nu}\,\half\,\pars{{1 \over t} - 1}^{-1/2}\pars{-\,{1 \over t^{2}}}\,\dd t =\half\,\mu^{-1/2}\int_{0}^{1}t^{\nu - 3/2}\pars{1 - t}^{-1/2}\,\dd t \\[3mm]&=\color{#00f}{\large\half\,\mu^{-1/2}\ \overbrace{{\rm B}\pars{\nu - \half,\half}}^{\mbox{Beta Function}}} \end{align}

\begin{align} \color{#00f}{\large\int_{-\infty}^{\infty}\pars{a^2 + x^{2}}^{-b}\,\dd x} &={2 \over \verts{a}^{2b}}\int_{0}^{\infty} \pars{1 + {1 \over a^{2}}\,x^{2}}^{-b}\,\dd x ={2 \over \verts{a}^{2b}}\,\half\,\verts{a}{\rm B}\pars{b - \half,\half} \\[3mm]&=\color{#00f}{\large{1 \over \verts{a}^{2b - 1}}\,{\rm B}\pars{b - \half,\half}} \end{align}

\begin{align} \color{#00f}{\large\int_{-\infty}^{\infty}\pars{a^2 + x^{2}}^{-2b}\,x^{2}\dd x} &={2 \over \verts{a}^{2b}}\int_{0}^{\infty} \pars{1 + {1 \over a^{2}}\,x^{2}}^{-2b}\,x^{2}\dd x \\[3mm]&=\left.{2 \over \verts{a}^{2b}}\pars{-\,{1 \over \nu}}\, \partiald{{\cal I}\pars{\mu,\nu}}{\mu} \right\vert_{\mu = {1 \over a^{2}},\nu = 2b - 1} \\[3mm]&={2 \over \verts{a}^{2b}}\pars{-\,{1 \over 2b - 1}} \pars{-\,{1 \over 4}\,\verts{a}^{3}}\,{\rm B}\pars{2b - {3 \over 2},\half} \\[3mm]&=\color{#00f}{\large% \half\,{1 \over \verts{a}^{2b - 3}}\,{1 \over 2b - 1}\, {\rm B}\pars{2b - {3 \over 2},\half}} \end{align}

Answer 3

Let us first evaluate $$I=\int_{\mathbb{R}} \dfrac{dx}{(x^2+a^2)^b}$$ Setting $x=\vert a \vert\tan(t)$, we get $$I = \int_{-\pi/2}^{\pi/2} \dfrac{\vert a \vert \sec^2(t)dt}{\vert a \vert^{2b} \sec^{2b}(t)} = \vert a \vert^{1-2b} \int_{-\pi/2}^{\pi/2} \cos^{2b-2}(t)dt = \vert a \vert^{1-2b}\beta(b-1/2,1/2)$$ where $\beta(x,y)$ is the beta function.

Mimic the same idea for the second integral as well and recall $$\beta(m+1,n+1) = 2\int_0^{\pi/2} \cos^{2m+1}(t) \sin^{2n+1}(t) dt$$