$\frac{1}{2 \pi i} \int_{a-i \infty}^{a+ i \infty} e^{st} \hat{f}(s) ds = \frac{2 e^{at}}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut du$

Question

The claim is from this paper: https://www.researchgate.net/publication/220668741_Numerical_Inversion_of_Laplace_Transforms_of_Probability_Distributions. The whole thing goes as:

And where $\hat{f}$ is the Laplace transfrom of $f$.

But I dont get the last line. Where does that come from? Function $f$ is assumed to be real, so I understand how imaginary parts cancel each other away, but the last line seem to assume $\Re(\hat{f}(a+iu))\cos ut = -\Im(\hat{f}(a+iu))\sin ut$ (!?), which kind of result I have never heard of and could not find from the references.

Thanks for help.

Answer 1

The result follows from monotone class theorem for functions Wikipedia. Let $\mathcal{H}$ be a set of functions $f$ which admits the condition $$ \int_{-\infty}^{\infty} \Re \hat{f}(a+ui) \cos(ut) du = -\int_{-\infty}^{\infty} \Im \hat{f}(a+ui) \sin(ut) du, $$ or equivalently $$ \int_{0}^{\infty} \int_{0}^{\infty}e^{-a \xi} \sin (u \xi) \cos(ut)f(\xi) d \xi du = \int_{0}^{\infty} \int_{0}^{\infty} e^{-a \xi} \sin(u \xi) \sin(ut)f(\xi) d\xi du. $$ Assuming $a>0$ one can verify the following cases:

(1) If $A \subset \mathbb{R}_+ \cup \{0\},$ then by evaluating the integrals one can conclude that $ f(x) = \mathbb{1}(x \in A) \in \mathcal{H}.$

(2) If $g,f \in \mathcal{H},$ then by linearity $g+f, cf \in \mathcal{H}$ for $ c \in \mathbb{R}.$

(3) If $f_n \in \mathcal{H}$ is a sequence of non-negative functions that increase to a bounded function $f,$ then by monotone convergence theorem $f \in \mathcal{H}.$

Consequently all bounded integrable functions on $\mathbb{R}_+ \cup \{0\}$ belongs to $\mathcal{H}.$

Answer 2

If $f$ is real valued and $a,u$ are real then $\hat{f}(a+iu) = \overline{\hat{f}(a-iu)}$ so that (assuming the Fourier inversion theorem)

$$e^{-at} f(t) = \frac{e^{-at}}{2 \pi i} \int_{a-i \infty}^{a+ i \infty} e^{st} \hat{f}(s) ds = \frac{1}{2\pi}\Re( \int_{-\infty}^\infty \hat{f}(a+iu))e^{iut}dt)\\= \frac{ 1}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut \ dt- \frac{ 1}{\pi} \int_0^{\infty} \Im(\hat{f}(a+iu)) \sin ut \ du$$ one term is even and the other one is odd so it must be $$ = \frac{e^{-at} f(t)+e^{at} f(-t)}{2} \quad -\quad\frac{e^{at} f(-t)-e^{-at} f(t)}{2}$$ whence for $t > 0$, if also $f$ vanishes on $t < 0$ it is $$f(t) = 2 \frac{e^{-at} f(t)+e^{at} f(-t)}{2} = \frac{ 2}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut \ dt$$