Let function F be defined as F(x) = _1^x e^tt,dt, x0 then the value of the integral _1^x …

Question

Let function $F$ be defined as $F(x) = \displaystyle\int_1^x \dfrac{e^t}{t}\,dt$, $x>0$ then the value of the integral $\displaystyle\int_1^x \dfrac{e^t}{t+a}\,dt$, where $a>0$, is:

Accepted Answer

Let $I = \int_1^x \dfrac{e^t}{t+a}\,dt$. Put $t+a=z \Rightarrow t=z-a$, $dt=dz$; limits $t=1 \Rightarrow z=1+a$, $t=x \Rightarrow z=x+a$

$I = \int_{1+a}^{x+a} \dfrac{e^{z-a}}{z}\,dz = e^{-a}\int_{1+a}^{x+a} \dfrac{e^z}{z}\,dz$

$= e^{-a}\left[\int_{1+a}^1 \dfrac{e^t}{t}\,dt + \int_1^{x+a}\dfrac{e^t}{t}\,dt\right] = e^{-a}[-F(1+a)+F(x+a)]$

$= e^{-a}[F(x+a)-F(1+a)]$

Solutions · 1