I want to evaluate the following integral in woflram alpha
int[x e^(-lambda*(x-v)),{x,v,infinity} ]
It doesn't work for some reason. It does work when instead of v, I put a number.
For example this works:
int[x e^(-lambda*(x-1)),{x,1,infinity} ]
Essentially there is no difference between the two.
Does someone know?
On
Mathematica works just fine:
Integrate[x Exp[-L (x - v)], {x, v, Infinity}]
gives the output
ConditionalExpression[(1 + L v)/L^2, Re[L] > 0]
We can also do the integral manually: with the choice $u = \lambda(x-v)$, $du = \lambda \, dx$, we obtain $$\begin{align*} \int_{x=v}^\infty x e^{-\lambda(x-v)} \, dx &= \int_{u=0}^\infty \frac{1}{\lambda} \left(\frac{u}{\lambda} + v\right) e^{-u} \, du \\ &= \frac{1}{\lambda^2} \int_{u=0}^\infty u e^{-u} \, du + \frac{v}{\lambda} \int_{u=0}^\infty e^{-u} \, du \\ &= \frac{1 + \lambda v}{\lambda^2}, \end{align*}$$ as claimed. For real-valued $\lambda$, this requires $ \lambda > 0$: if $\lambda \le 0$, the integral fails to converge.
Hint
You are more clever than Wolfram Alpha. So use your knowledge for your problem (only you know at least the sign of $\lambda$).
Since $$x e^{-\lambda (x-v)}=e^{\lambda v} x e^{- \lambda x}$$ then $$I=\int_v^{\infty}x e^{-\lambda (x-v)}dx=e^{\lambda v}\int_v^{\infty} x e^{- \lambda x}dx$$ Now, change variable $x=\frac y{\lambda}$. So $$I=\frac{e^{\lambda v}}{\lambda}\int_1^{\infty} y e^{- y}dy$$ Now, use Wolfram Alpha ... or, much better, compute the antiderivative (it is simple).