Integration possible using DiracDelta or in another way?

Question

Could MMA solve analytically this integral by using a Dirac delta function or in another way? $$ f(x)=\int_{-\infty}^{\infty}\left(\frac{1}{3 t^2+1} e^\frac{-t^2+i t}{3 t^2+1}\right)e^{itx}{\rm d}t\tag{1} $$

Integrate[1/(3*t^2+1)*Exp[(-t^2+I*t)/(3*t^2+1)]*Exp[I*t*x],{t,-\[Infinity],\[Infinity]}]

Below is a plot of $f(x)$. What is the expression for $f(x)$ after solving the integral? In case it is helpful, the first moments $\mu_n=\int_{-\infty}^{\infty}x^n f(x){\rm d}x$ are $\mu_0=2\pi,\mu_1=-2\pi,\mu_2=18\pi,\mu_3=-86\pi,\mu_4=986\pi,\mu_5=-9282\pi,\mu_6=133322\pi$.

Solved related problem using DiracDelta

MMA could solve the similar integral in eq.(2) $$ I=\int_{-\infty}^{\infty}\left(\frac{1}{3 t^2+1} e^\frac{-t^2+i t}{3 t^2+1}\right)\color{red}{ \int_{-\infty}^{\infty}x}e^{itx}\color{red}{{\rm d}x}{\rm d}t \tag{2} $$ where deviations to eq.(1) are colored (see also this post). Using the Dirac delta function $\delta(t)$ $$ \delta(t)=\frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{i t x}{\rm d}x \tag{3} $$ $$\delta'(t)=\frac{i}{2\pi}\int_{-\infty}^{\infty} xe^{i t x}{\rm d}x \tag{4} $$ MMA solved instantly:

$$I=-2\pi i\int_{-\infty}^{\infty}\frac{1}{3t^2+1}e^{\frac{-t^2+it}{3t^2+1}}\delta'(t){\rm d}t\\=-2\pi i\left(\left[\frac{1}{3t^2+1}e^{\frac{-t^2+it}{3t^2+1}}\delta(t)\right]_{-\infty}^{\infty}-\frac{{\rm d}}{{\rm d}t}\left(\frac{1}{3t^2+1}e^{\frac{-t^2+it}{3t^2+1}}\right)_{t=0}\right)\\=-2\pi i(0-i)=-2\pi \tag{5}$$

Integrate[-2*Pi*I /(3*t^2+1)*Exp[(-t^2+I*t)/(3*t^2+1)]*D[DiracDelta[t],t],{t,-\[Infinity],\[Infinity]}]
(*-2 Pi*)

MMA 13

Edit note

In a previous version in an exponent was wrongly written $3t^2+t$ instead of $3t^2+1$.

Answer 1

To summarize the discussion:

Integrate[1/(3*t^2+1)*Exp[(t^2+I*t)/(3*t^2+1)]*Exp[I*t*x]
{t,-\[Infinity],\[Infinity]}]

represents a Fourier integral. The DiracDelta manipulations are also stated in terms of Fourier integrals. Mathematica has special methods that it uses in this domain which can sometimes perform symbolic integrations that are not possible otherwise. However:

FourierTransform[-2*Pi*I/(3*t^2 + 1)*Exp[(-t^2 + I*t)/(3*t^2 + 1)],
 t,x]

returns unevaluated in this case, suggesting that those special methods do not suffice.

Answer 2

An approximate answer to the edited question can be derived as follows, expanding Exp[(-t^2+I*t)/(3*t^2+1)] into a series.

Total[Table[FourierTransform[1/(3*t^2 + 1)*((-t^2 + I*t)/(3*t^2 + 1))^n/n!,t, x], {n, 0, 20}]];
Plot[Sqrt[2*Pi]*%, {x, -8, 8}]

Edit. Plot[Sqrt[2*Pi]*%, {x, -8, 8}] instead of Plot[%, {x, -8, 8}].

Answer 3

As noticed in @John Doty comment, the integral under consideration is a Fourier transform of $\frac{1}{3 t^2+1} e^\frac{-t^2+i t}{3 t^2+t} $ up to a multiplier. The necessary condition of its existence (see e.g. Encyclopedia of Mathematics) is the convergence of the integral $$ \int_{-\infty}^\infty\left|\frac{1}{3 t^2+1} e^\frac{-t^2+i t}{3 t^2+t}\right|\,dt.$$ There is a non-integrable singularity of the integrand at $t=-\frac 1 3$ as the Mathematica commands

as = Normal[Series[Abs[1/(3 t^2 + 1)*Exp[(-t^2 + I*t)/(3 t^2 + t)]], {t,-1/3, 2}]]

E^Re[(I t - t^2)/(t + 3 t^2)]/Abs[1 + 3 t^2]

Plot[as, {t, -0.5, 0}]

NIntegrate[as, {t, -0.5, 0}]

3.632573614239987*10^477NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in t near {t} = {-0.333016}. NIntegrate obtained 3.63257361423998710^477+0. I and 3.580550483737237`15.954589770191005^477 for the integral and error estimates.

report. Therefore, the integral under consideration does not exist.