I encounter a problem in which I would need to deal with the folloing definite integral $$I(t)=\int_{0}^{\infty} \mathrm dp \frac{p^2}{\omega^5} \sin^2\left(\frac{\omega t}{2}\right)$$ in which $$\omega=\sqrt{m^2+p^2}$$ where m is just some positive constant, t is also a parameter. But I would like to obtain the value of $I(t)$ when t is small. My approach to this problem is the following $$I(t)=I(0)+I(0)'t+\frac{I(0)''}{2}t^2...$$ $$\frac{dI}{dt}|_{t=0}=0$$ However $$I(0)''\propto \int dp \frac{p^2}{\omega^3}$$ which diverge like $\ln(p)$ Can anyone help me to find the expansion of I(t) in terms of t?
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2The integral diverges but it exists as a tempered distribution, as would all future derivatives since they grow at most polynomially. The trick is to choose a suitable Schwartz function $f_\lambda$ such that the integrand $f_\lambda \cdot p^2\omega^n$ is easier to integrate and $f_\lambda\to1$ as $\lambda\to0^+$ – Ninad Munshi Apr 08 '22 at 15:18
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@NinadMunshi Hi, I am a bit confused. Could you write it out explicitly? I would be more than grateful for your help. – Tan Tixuan Apr 08 '22 at 15:28
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What are you confused about? I cannot elaborate without knowing specifically what to elaborate on – Ninad Munshi Apr 08 '22 at 15:30
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At any $t\neq 0$ integral is bounded by 0 from below and by $\int_0^1 \frac{p^2}{m^5}dp + \int_1^\infty \frac{p^2}{p^5}dp$ from above. One cannot be sure that $I(t)$ is continuous at $t=0$. – Ivan Kaznacheyeu Apr 08 '22 at 16:51
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@IvanKaznacheyeu I asked someone else, they say that this integral is C1 but not C2. And this can be proved by "dominant convergence theorem". But I am not particularly sure of the proof – Tan Tixuan Apr 09 '22 at 04:41
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Note that $I(t) = I_1(mt)/m^2$, where $I_1$ is the value of the integral for $m=1$. So we only need consider the case $m=1$. – eyeballfrog Apr 09 '22 at 16:04
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@eyeballfrog. No, you still have $m$ in the sine. Cheers – Claude Leibovici Apr 11 '22 at 08:10
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@ClaudeLeibovici Don't forget the substitution you have to do for rescaling $p$: $mt\sqrt{1 + u^2} = t\sqrt{m^2 + (mu)^2} = t\sqrt{m^2 + p^2}$ – eyeballfrog Apr 11 '22 at 11:13
4 Answers
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@ClaudeLeibovici got a closed form of the asymptotics term $\sim t^2\ln t$. It would be useful, though, to get the closed form of the second term $\sim t^2$. Given that logarithm is a slowly growing function, the term $\sim t^2$ has a significant contribution at any reasonable $t$. The next asymptotics terms have a contribution $\sim t^4\ln t$, so they are much smaller. We will use the heuristic approach to get the answer.
Let's denote the desired integral as $$\boxed{\,I(t)=\int_0^\infty\frac{p^2}{(m^2+p^2)^\frac{5}{2}}\sin^2\frac{\sqrt{m^2+p^2}\,t}{2}dp\,}=\int_0^\infty\frac{p^2}{(m^2+p^2)^\frac{5}{2}}\Big(\frac{1}{2}-\frac{\cos(\sqrt{m^2+p^2}\,t)}{2}\Big)dp$$
$$=\frac{1}{2}\int_0^\infty\frac{p^2}{(m^2+p^2)^\frac{5}{2}}dp-\frac{1}{2}\int_0^\infty\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{3}{2}}dp+\frac{m^2}{2}\int_0^\infty\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{5}{2}}dp$$ Decomposing $\cos$ at small $t$ in the third term $$\frac{m^2}{2}\int_0^\infty\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{5}{2}}dp=\frac{m^2}{2}\int_0^\infty\frac{dp}{(m^2+p^2)^\frac{5}{2}}-\frac{t^2}{4}\int_0^\infty\frac{dp}{(m^2+p^2)^\frac{3}{2}}+o(t^2)$$ $$I(t)=-\frac{1}{2}\int_0^\infty\bigg(\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{3}{2}}-\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{5}{2}}\bigg|_{t=0}\bigg)dp-\frac{(mt)^2}{4}\int_0^\infty\frac{dp}{(m^2+p^2)^\frac{3}{2}}+o(t^2)$$ The evaluation of the second integral is straightforward: $$I_2=-\frac{(mt)^2}{4}\int_0^\infty\frac{dp}{(m^2+p^2)^\frac{3}{2}}=-\frac{t^2}{4}$$ Now, we have to find the asymptotics at $t\to0$ of $$I_1=-\frac{1}{2}\int_0^\infty\frac{\cos(\sqrt{m^2+p^2}\,t)}{(m^2+p^2)^\frac{3}{2}}dp=-\frac{1}{2m^2}\int_0^\infty\frac{\cos(mt\,\sqrt{x^2+1})}{(x^2+1)^\frac{3}{2}}dx$$ Making the substitution $x=\sinh s$ $$=-\frac{1}{2m^2}\int_0^\infty\frac{\cos(mt\,\cosh s)}{\cosh^2s}ds=-\frac{1}{2m^2}J(\alpha=mt)$$ where $J(\alpha)=\int_0^\infty\frac{\cos(\alpha\cosh s)}{\cosh ^2s}ds$. Taking the second derivative of $J(\alpha)$ with respect to $\alpha$ $$J''(\alpha)=-\int_0^\infty\cos(\alpha\cosh s)ds=\frac{\pi}{2}Y_0(\alpha)$$ where $$Y_0(z) = -\frac{2}{\pi}\int_0^\infty\cos(z\cosh s)ds=\frac{2}{\pi}\sum_{r=0}^\infty\frac{(-1)^r(\frac{z}{2})^{2r}}{(r!)^2}\Big(\ln\frac{z}{2}-\psi(r+1)\Big)$$ is the Bessel function of the second kind (here (10.9.9) and, for example, here ( page 24)); $\,\psi(r)$ is digamma-function. At small $z$ $$Y_0(z)=\frac{2}{\pi}\Big(\ln\frac{z}{2}-\psi(1)+O(z^2\ln z)\Big)$$ it is easy to see that all other terms ($\sim z^2\ln z$ or $z^2$) will contribute to the term $\sim z^4\ln z$ or $z^4$ of the asymptotics and can be dropped. Integrating $$J(\alpha)=\int^\alpha ds\int^s J''(z)dz=C_2+C_1\alpha+\frac{\alpha^2\ln\alpha}{2}-\frac{\alpha^2}{2}\Big(\frac{3}{2}+\psi(1)+\ln2\Big)+o(\alpha^2\ln\alpha)$$ where $C_1, C_2$ - some constants. It is easy to check that $C_1=0$ (the asymptotics does not contain the term $\sim \alpha$). We also have to extract $J(0)=C_2$ $$I(t)=\big(I_1(t)-I_1(0)\big)+I_2=\boxed{\,-\frac{1}{4}t^2\ln(mt)+\frac{t^2}{4}\Big(\frac{1}{2}-\gamma+\ln 2\Big)+o(t^2\ln t)\,}$$ We can present the answer in a more convenient form: $$m^2\,I(mt)=\int_0^\infty\frac{x^2}{(x^2+1)^\frac{5}{2}}\sin^2\frac{mt\sqrt{x^2+1}}{2}dx$$ $$=-\frac{(mt)^2}{4}\ln(mt)+\frac{(mt)^2}{4}\Big(\frac{1}{2}-\gamma+\ln2\Big)+O\Big((mt)^4\ln(mt)\Big)$$ Quick check: for $mt=0.1$ WolframAlpha gives the integral value $$m^2\,I(mt=0.1)=0.007301501...$$ Our approximation gives $$\bigg(-\frac{(mt)^2}{4}\ln(mt)+\frac{(mt)^2}{4}\Big(\frac{1}{2}-\gamma+\ln2\Big)\bigg)\bigg|_{mt=0.1}=0.00729629...$$
Svyatoslav
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@Claude Leibovici , thank you! Your solution inspired me to find the second asymptotics term :) – Svyatoslav Apr 11 '22 at 15:09
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$$I=\int\frac{p^2 }{\left(m^2+p^2\right)^{5/2}}\sin ^2\left(\frac{t}{2} \sqrt{m^2+p^2}\right)\,dp$$ $$\frac{t}{2} \sqrt{m^2+p^2}=x\implies p=\frac{\sqrt{4 x^2-m^2 t^2}}{t}\implies dp=\frac{4 x}{t \sqrt{4 x^2-m^2 t^2}}\,dx$$ $$I=\frac {t^2} 8 \int \frac{ \sqrt{4 x^2-m^2 t^2}}{ x^4}\,\sin ^2(x)\,dx $$ Now, expanding $\sqrt{4 x^2-m^2 t^2}$ as series around $t=0$ we face the problem of $$I=\sum_{n=1}^\infty J_n$$ $$J_n=(-1)^{n-1} \,2^{-2 n} \binom{\frac{1}{2}}{n-1} m^{2( n-1)} t^{2 n}\int_{\frac {mt}2}^\infty x^{-(2 n+1)} \sin ^2(x)\,dx$$ where all antiderivatives are known (they are $0$ at $\infty$).
Expanded as series to $O(t^4)$, we the have $$\frac I{t^2}=-\frac{1}{4} (\log (mt)+\gamma )+\Bigg[\frac 38-\sum_{n=2}^\infty a_n \Bigg]+O(t^2)$$ and the first $a_n$ form the sequence $$\left\{\frac{1}{16},\frac{1}{128},\frac{1}{384},\frac{5}{4096},\frac{7}{10240},\frac{7}{16384},\frac{33}{114688},\frac{429}{2097152},\frac{715}{4718592},\frac{2431 }{20971520},\cdots\right\}$$ Using these terms only $$\frac 38-\sum_{n=2}^{11} a_n =\frac{395030009}{1321205760}$$
Now, some results (using $m=1$ $$\left( \begin{array}{ccc} t & \text{estimation} & \text{solution} \\ 0.01 & 0.000130598 & 0.000130529 \\ 0.02 & 0.000453078 & 0.000452807 \\ 0.03 & 0.000928195 & 0.000927614 \\ 0.04 & 0.001535051 & 0.001534083 \\ 0.05 & 0.002259053 & 0.002257664 \\ 0.06 & 0.003088947 & 0.003087157 \\ 0.07 & 0.004015565 & 0.004013456 \\ 0.08 & 0.005031170 & 0.005028293 \\ 0.09 & 0.006129064 & 0.006126850 \\ 0.10 & 0.007303344 & 0.007301514 \\ 0.15 & 0.014151783 & 0.014160180 \\ 0.20 & 0.022281904 & 0.022325476 \\ 0.25 & 0.031328857 & 0.031451016 \\ 0.30 & 0.041011319 & 0.041277079 \\ 0.35 & 0.051100098 & 0.051599323 \\ 0.40 & 0.061401730 & 0.062252075 \\ 0.45 & 0.071748798 & 0.073098053 \\ 0.50 & 0.081993730 & 0.084022233 \end{array} \right)$$
The last point to be checked is that this shows that, for a given $t$ $$I=\alpha -\beta \, \log(m)$$
For $t=0.1$, the results are $$\left( \begin{array}{cc} m & I \\ 1 & 0.0073015138 \\ 2 & 0.0055813716 \\ 3 & 0.0045863288 \\ 4 & 0.0038907431 \\ 5 & 0.0033608646 \\ 6 & 0.0029369547 \\ 7 & 0.0025869277 \\ 8 & 0.0022916176 \\ 9 & 0.0020385018 \\ 10 & 0.0018190103 \end{array} \right)$$
A linear regression gives, with $R^2=0.999900$,
$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 0.0072411 & 0.0000362 & \{0.0071555,0.0073268\} \\ \beta & 0.0023846 & 0.0000218 & \{0.0023331,0.0024361\} \\ \end{array}$$ while the above give $\beta=0.0025$
Update
Computing the various $J_n$ before expansions, we have $$I=\sum_{k=1}^\infty \frac {m^{2(k-1)}} {a_k} (\log (mt)+\gamma- b_k )\,t^{2k}$$ $$I=\frac 1 {m^2}\sum_{k=1}^\infty \frac {1} {a_k} (\log (mt)+\gamma- b_k )\,(mt)^{2k}$$ where the first $a_k$ are $$\left\{-4,-96,11520,-1290240,185794560,-35035545600,\cdots\right\}$$ The $b_k$ being rational numbers with huge numerators and denominators, they were converted to decimals and rationalized (for an arror lower than $10^{-10}$) $$\left\{ \frac{1400566}{1173575} ,\frac{211825}{64661},\frac{152687}{52816},\frac{729301}{243927},\frac{136163}{43598},\frac{614252}{188927},\cdots\right\}$$
For $m=1$ and $t=\frac 12$, this gives $I=0.0840385$ while the exact value is $0.0840222$.
Claude Leibovici
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Nice solution (+1)! The leading asymptotics term ($-\frac{1}{4}t^2\ln t$) has a closed form. – Svyatoslav Apr 09 '22 at 16:52
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With Mathematica I have:
$$\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{1}{2} \sqrt{m^2+p^2} t\right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\int_0^{\infty } \left(\frac{p^2}{2 \left(m^2+p^2\right)^{5/2}}-\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right) \, dp=\\\frac{1}{6 m^2}-\frac{\pi G_{1,3}^{2,0}\left(\frac{m t}{2},\frac{1}{2}| \begin{array}{c} \frac{5}{2} \\ 0,1,\frac{1}{2} \\ \end{array} \right)}{8 m^2}$$
for: $t>0$
MMA code:
Integrate[(p^2 Sin[1/2 Sqrt[m^2 + p^2] t]^2)/(m^2 + p^2)^( 5/2), {p, 0, \[Infinity]}] == 1/(6 m^2) - (\[Pi] MeijerG[{{}, {5/2}}, {{0, 1}, {1/2}}, (m t)/2, 1/ 2])/(8 m^2)
Using MellinTransfrom:
$\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_t\left[\frac{p^2 \cos \left(\sqrt{m^2+p^2} t\right)}{2 \left(m^2+p^2\right)^{5/2}}\right](s) \, dp\right](t)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{1}{2} p^2 \left(m^2+p^2\right)^{-\frac{5}{2}-\frac{s}{2}} \cos \left(\frac{\pi s}{2}\right) \Gamma (s) \, dp\right](t)=\\\mathcal{M}_s^{-1}\left[\frac{m^{-2-s} \sqrt{\pi } \cos \left(\frac{\pi s}{2}\right) \Gamma \left(1+\frac{s}{2}\right) \Gamma (s)}{8 \Gamma \left(\frac{5+s}{2}\right)}\right](t)=\\\frac{\pi G_{1,3}^{2,0}\left(\frac{m t}{2},\frac{1}{2}| \begin{array}{c} \frac{5}{2} \\ 0,1,\frac{1}{2} \\ \end{array} \right)}{8 m^2}$
Mariusz Iwaniuk
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I prefer to add another answer since based on a different approach.
$$I=\int_0^{\infty } \frac{p^2 \sin ^2\left(\frac{t}{2} \sqrt{m^2+p^2} \right)}{\left(m^2+p^2\right)^{5/2}} \, dp=\frac12\int_0^{\infty } \frac{p^2}{ \left(m^2+p^2\right)^{5/2}}\, dp-\frac12\int_0^{\infty } \frac{p^2 \cos \left(t\sqrt{m^2+p^2} \right)}{ \left(m^2+p^2\right)^{5/2}} \, dp$$
Let $p=m q$ $$2m ^2\,I=\int_0^{\infty }\frac{q^2}{\left(q^2+1\right)^{5/2}}\,dq-\int_0^{\infty }\frac{q^2 \cos \left(m t\sqrt{q^2+1} \right)}{\left(q^2+1\right)^{5/2}}\,dq$$ $$\frac 13-2m^2\,I=\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1} \right)}{\left(q^2+1\right)^{5/2}}\,dq$$ $$mt \sqrt{q^2+1}=x \implies q=\frac{\sqrt{x^2-m^2 t^2}}{m t}\implies dq=\frac{x}{m t \sqrt{x^2-m^2 t^2}}\,dx$$ $$\int_0^{\infty }\frac{q^2 \cos \left(mt \sqrt{q^2+1} \right)}{\left(q^2+1\right)^{5/2}}\,dq=(mt)^2 \int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$ $$\frac{1-6m^2\,I}{(mt)^2}=\int_{mt}^\infty \sqrt{x^2-m^2 t^2}\,\,\frac{\cos (x) }{x^4}\,dx$$ $$\sqrt{x^2-m^2 t^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\, x^{1-2 n}$$ $$\frac{1-6m^2\,I}{(mt)^2}=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n} (m t)^{2 n}\int_{mt}^\infty \frac {\cos (x)}{x^{2 n+3} }\,dx$$ $$2 \left(6 m^2\,I-1\right)=\sum_{n=0}^\infty (-1)^n \binom{\frac{1}{2}}{n}\Big[E_{2 n+3}(+i m t)+E_{2 n+3}(-i m t)\Big]$$
Claude Leibovici
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