integration of a log-normal random variable Let X be a log-normal disributed random variable, $\forall{a}\in\mathbb{R}$ compute: $$\int_{\mathbb{R}}|a+X|dP.$$
I get that when $a\ge0$, since $X$'s support is $[0,\infty)$, holds:
$$\int_{\mathbb{R}}|a+X|dP=\int_{\mathbb{R}}(a+X)dP=a+E[X].$$
When $a<0$, is there a way of computing the integral without using the pdf like I did before?

Note $|a+X|=(a+X)^+-(a+X)^-=\max(a+X,0)-\min(a+X,0)$. If $a<0$ set $b:=-a,\,b>0$ and we have
$$|X-b|=(X-b)^+-(X-b)^-=\max(X-b,0)-\min(X-b,0)$$
The expectation $E[(X-b)^+]$ is given in closed form by the Black-Scholes formula for call options. On the other hand:
$$E[(X-b)^-]=-E[(b-X)^+]$$
which is again given by the Black-Scholes formula of put options.
