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I want to express it in the form

$$\text{expr=}\int_{-\infty}^{\infty}C_1\exp\left(-t^2\right)f(t)dt$$, i.e., the integrand is a bounded smooth function $f(t)$ times $\exp(-t^2)$ and the domain of integration is the whole line.

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Define the integrand with the measure as

integrand = c*Exp[-m*x^2/ω^m]*1/(Sqrt[2*π]*λ*ω)*Exp[-(Log[B] - μ)^2/(2*λ^2)] Dt[B, Constants -> {λ, m, x, μ, ω}];

where we use Dt[t] to represent the measure, which we will transform using the substitution:

subs = Assuming[{μ > 0, λ > 0, B > 0, t ∈ Reals}, First@Solve[t == (Log[B] - μ)/(Sqrt[2] λ), B, Reals]]
(* {B -> E^((2 t λ + Sqrt[2] μ)/Sqrt[2])} *)

The transformation is done via

transformedIntegrand = integrand /. subs /. Log[Exp[x_]] :> x // Simplify
(* (c E^(-t^2 + Sqrt[2] t λ + μ - m x^2 ω^-m) Dt[t, Constants -> {m, x, λ, μ, ω}])/(Sqrt[π] ω) *)

where we have simplified the expression by hand to some extent.

Finally, we integrate:

Integrate[transformedIntegrand /. HoldPattern[Dt[a__]] :> 1, {t, -∞, ∞}]
(* (c E^(λ^2/2 + μ - m x^2 ω^-m))/ω *)

(We have to set the measure back to 1, because obviously in Mathematica you don't specify the measure explicitly.)

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  • $\begingroup$ Thanks for you answer. Sorry, I mistype my MMA code...Please check your answer...$B$ in fact should be $\omega$. There is no $B$ in the expression.. $\endgroup$ – Srestha Narayanan May 5 '16 at 7:00

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