If you expand the integrand to ReIm

reim=(f[x] g[x])/(\[Lambda] + I x) // ComplexExpand[ReIm[#]] & // Simplify
(*{(2 (1 + x^2) \[Lambda])/(\[Pi] (1 + 4 x^2)(x^2 + \[Lambda]^2)), -((2 (x + x^3))/(\[Pi] (1 + 4 x^2)(x^2 + \[Lambda]^2)))}*)

integration gives (Mathematica v12.2)

Integrate[reim,{x, -Infinity, Infinity}]

$\left\{\fbox{$\frac{\lambda +2}{2 \lambda +1}\text{ if }\Re(\lambda )>0$},\int_{-\infty }^{\infty } -\frac{2 \left(x^3+x\right)}{\pi \left(4 x^2+1\right) \left(\lambda ^2+x^2\right)} \, dx\right\}$

That means only integration over real part converges!

If you expand the integrand to ReIm

reim=(f[x] g[x])/(\[Lambda] + I x) // ComplexExpand[ReIm[#]] & // Simplify
(*{(2 (1 + x^2) \[Lambda])/(\[Pi] (1 + 4 x^2)(x^2 + \[Lambda]^2)), -((2 (x + x^3))/(\[Pi] (1 + 4 x^2)(x^2 + \[Lambda]^2)))}*)

integration gives (Mathematica v12.2)

Integrate[reim, {x, -Infinity, Infinity}]

$\left\{\fbox{$\frac{\lambda +2}{2 \lambda +1}\text{ if }\Re(\lambda )>0$},\int_{-\infty }^{\infty } -\frac{2 \left(x^3+x\right)}{\pi \left(4 x^2+1\right) \left(\lambda ^2+x^2\right)} \, dx\right\}$

That means only integration over real part converges.