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$f(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ and $f(x) = 0$ for $x <0$

Here is what I got:

$$\phi_x(t)=\mathbb{E} [e^{itX}]$$

$$=\int_\mathbb{R} e^{itx} f_X (x) dx=\int_0^\infty (\cos x + i\sin x)t\lambda e^{-\lambda x} dx$$...(a little bit suspicious here changing from $\mathbb{R}$ to $[0,\infty]$

Then split the integral into two parts, the first part without imaginary number is $$\int_0^\infty t\lambda \cos x e^{-\lambda x} dx = \lim_{k\to \infty}\int_0^k t \lambda \cos x e^{-\lambda x} dx$$

Trusting WolframAlpha, I have: $$t \lim_{k\to \infty} \frac{\lambda e^{-\lambda k}(\lambda e^{\lambda k} - \lambda \cos(k) + \sin (k))}{\lambda^2 + 1}$$ But then I'm not sure how to deal with this limit especially the sin, cos part. Any help? Thanks!

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1 Answer 1

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Why not do it directly: $$ \int e^{itx} \lambda e^{-\lambda x} dx = \lambda \int e^{(it-\lambda)x}dx = \frac{\lambda}{it-\lambda} e^{(it-\lambda)x} $$ and then you do arithmetic

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