How to calculate such integral? Treating $i$ as constant obviously doesn't work.
$$\int_{0}^{\infty} \exp[(it-\lambda)x] dx$$ $t\ \in\mathbb R, \lambda\ > 0$.
P.S. Above integral allows us to calculate characteristic function of exponential distribution.
Try to integrate for the real part and the imaginary part separately:
$$\int u(x)+iv(x)\ dx\ =\ \int u(x)\,dx+i\int v(x)\,dx$$
where both $\,u(x),\,v(x)\,$ are real-valued functions
For this question, we have
\begin{align*} &\int_0^\infty\exp[(it-\lambda)x]dx\\ =\ &\int_0^\infty\exp[i(tx)]\cdot\exp(-\lambda x)\ dx\\ =\ &\int_0^\infty\left[\cos(tx)+i\cdot\sin(tx)\right]\cdot\exp(-\lambda x)\ dx\\ =\ &\int_0^\infty\exp(-\lambda x)\cdot\cos(tx)\ dx+i\int_0^\infty\cdot\exp(-\lambda x)\cdot\sin(tx)\ dx\\ \end{align*}
Then you can integrate by parts to finish this question:
$$\int_0^\infty\exp(-\lambda x)\cdot\cos(tx)\ dx=\left[\exp(-\lambda x)\frac{\sin(tx)}{t}\right]_0^\infty-\int_0^\infty-\frac\lambda t\cdot\exp(-\lambda x)\sin(tx)\ dx$$
$$=-\left[\frac\lambda{t^2}\cdot\exp(-\lambda x)\cos(tx)\right]_0^\infty-\frac{\lambda^2}{t^2}\int_0^\infty\exp(-\lambda x)\cos(tx)\ dx$$
Solve the equation, we have
$$\int_0^\infty\exp(-\lambda x)\cdot\cos(tx)\ dx\ =\ \frac\lambda{t^2}\ \left/\frac{}{}\right.\left(1+\frac{\lambda^2}{t^2}\right)\ =\ \frac\lambda{\lambda^2+t^2}$$
Similarly,
$$\int_0^\infty\exp(-\lambda x)\cdot\sin(tx)\ dx\ =\ \frac t{\lambda^2+t^2}$$
Thus,
$$\int_0^\infty\exp[(it-\lambda)x]dx=\frac{\lambda+it}{\lambda^2+t^2}$$