Characteristic function of the mixture of exponential and normal distributions

Question

I've been calculating the characteristic function of a random variable with distribution $\frac{2}{3}E(\frac{1}{3}) + \frac{1}{3}N(6, 3)$.

\begin{align} \phi(t) &= \int_{-\infty}^{0} e^{-itx}\frac{1}{3}\frac{1}{\sqrt{3}\sqrt{2\pi}}\exp\bigg({-\frac{1}{2}\frac{(x - 6)^2}{3}}\bigg) \mathrm{d}x + \int_0^{\infty} e^{-itx}\bigg(\frac{2}{9}e^{-x / 3} + \frac{1}{3}\frac{1}{\sqrt{3}\sqrt{2\pi}}\exp\bigg(-\frac{1}{2}\frac{(x - 6)^2}{3}\bigg)\bigg) \mathrm{d}x \\ &= \frac{1}{3^{3/2}\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-itx}\exp\bigg(-\frac{1}{2}\frac{(x - 6)^2}{3}\bigg) \mathrm{d}x + \frac{2}{9} . \int_0^{\infty} e^{-itx}\frac{e^{-x}}{3} \mathrm{d}x. \end{align}

For the first integral \begin{align} \int e^{-itx}\exp\bigg(-\frac{1}{2}\frac{(x - 6)^2}{3}\bigg) \mathrm{d}x &= \int \exp\bigg(-\frac{1}{6}\bigg(x^2 - 6(2 - it)x + 36\bigg)\bigg) \mathrm{d}x \\ &= \int \exp\bigg(-\frac{1}{2}\bigg(\bigg(\frac{x - (6 - 3it)}{\sqrt{3}}\bigg)^2 + 12it + 3t^2\bigg)\bigg) dx \qquad\text{by completing the square} \\ &= \sqrt{3}\exp\bigg(-\frac{1}{2}t(12i + 3t)\bigg)\int \exp\bigg(-\frac{1}{2}y^2\bigg) \mathrm{d}y \qquad\text{where }y = \frac{x - (6- 3it)}{\sqrt{3}} \\ &= \sqrt{3}\exp\bigg(-\frac{1}{2}t(12i + 3t)\bigg)\sqrt{2\pi}. \end{align}

For the second integral, \begin{align} \int_0^{\infty} e^{-itx}e^{-x / 3} \mathrm{d}x &= \int_0^\infty \exp\bigg(\bigg(-it - \frac{1}{3}\bigg)x\bigg) \mathrm{d}x \\ &= \frac{3}{3it + 1}. \end{align}

$$ \phi(t) = \frac{1}{3}\exp\bigg(-\frac{1}{2}t(12i + 3t)\bigg) + \frac{2}{3(3it + 1)}. $$

Did I get the computation right?

Answer 1

No. I should have used $e^{itx}$.

\begin{align} \int e^{itx}\exp\bigg(-\frac{1}{2}\frac{(x - 6)^2}{3}\bigg) \mathrm{d}x &= \int \exp\bigg(-\frac{1}{6}\bigg(x^2 - 6(2 + it)x + 36\bigg)\bigg) \mathrm{d}x \\ &= \int \exp\bigg(-\frac{1}{2}\bigg(\bigg(\frac{x - (6 + 3it)}{\sqrt{3}}\bigg)^2 - 12it + 3t^2\bigg)\bigg) dx \\ &= \sqrt{3}\exp\bigg(-\frac{1}{2}t(-12i + 3t)\bigg)\sqrt{2\pi}. \\ \\ \int_0^{\infty} e^{itx}e^{-x / 3} \mathrm{d}x &= \int_0^\infty \exp\bigg(\bigg(it - \frac{1}{3}\bigg)x\bigg) \mathrm{d}x \\ &= \frac{3}{1 - 3it}.\\ \\ \phi(t) &= \frac{1}{3}\exp\bigg(-\frac{1}{2}t(-12i + 3t)\bigg) + \frac{2}{3(1 - 3it)}. \end{align}