# Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$

I am stuck trying to evaluate or simplify this integrale :

$$I_{a,b,c} = \int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$$

with $a,b,c \in \mathbb{R}_+^*$. and $Ei(x) =\int_{-\infty}^{x} \frac{\exp(t)}{t} \mathbb{d}t$ : The Exponential integral function

I already found this result: $$\int\exp(-bx) \ Ei(x) = \frac{1}{b} \left[Ei((1-b)x)-\exp(-bx)\ Ei(x) \right]$$ but I can't see if it useful.
Any Hint ?

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How about first doing the special case $\int_1^\infty \mathrm{Ei}(x) dx/(x+1)$...??? – GEdgar Jun 1 '12 at 21:48
@GEdgar in that case the integral does not converge. – jack Jun 1 '12 at 23:47
I should ask for special case $\int \mathrm{Ei}(x)dx/(x+1)$, since the original is asking for an indefinite integral of $\exp(-bx)\mathrm{Ei}(x)/(x+c)$. – GEdgar Jun 2 '12 at 0:30
@jack:Have a look to E1-Transform. As if your integral is related to that transform. Maybe the link can be helpful to you. m-hikari.com/ijcms-password2008/13-16-2008/… – Mathlover Jun 2 '12 at 7:52
@jack: $\text{Ei}(x)$ should not be $\int_{-\infty}^x\dfrac{e^t}{t}dt$ when $x>0$ and should be for example $\gamma+\ln x+\int_0^x\dfrac{e^t-1}{t}dt$ instead, as $\int_{-\infty}^x\dfrac{e^t}{t}dt$ diverges when $x>0$ . – doraemonpaul Aug 5 '13 at 1:44

By using the expression $\text{Ei}(x)=\gamma+\ln x+\int_0^x\dfrac{e^t-1}{t}dt$ mentioned in http://people.math.sfu.ca/~cbm/aands/page_230.htm,

$\int_a^\infty\dfrac{e^{-bx}\text{Ei}(x)}{x+c}dx$

$=\int_a^\infty\dfrac{e^{-bx}}{x+c}\left(\gamma+\ln x+\int_0^x\dfrac{e^t-1}{t}dt\right)~dx$

$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}\ln x}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_0^x\dfrac{e^t-1}{t}dt~dx$

$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_1^x\dfrac{1}{t}dt~dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_0^x\dfrac{e^t-1}{t}dt~dx$

$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\int_1^x\dfrac{e^{-bx}}{t(x+c)}dt~dx+\int_a^\infty\int_0^x\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dt~dx$

$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_1^a\int_a^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_0^a\int_a^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt$

$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_1^a\int_a^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_0^a\int_a^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{e^te^{-bx}}{t(x+c)}dx~dt$

$=\gamma\int_{a+c}^\infty\dfrac{e^{-b(x-c)}}{x}dx+\int_1^a\int_{a+c}^\infty\dfrac{e^{-b(x-c)}}{tx}dx~dt+\int_0^a\int_{a+c}^\infty\dfrac{(e^t-1)e^{-b(x-c)}}{tx}dx~dt+\int_a^\infty\int_{t+c}^\infty\dfrac{e^te^{-b(x-c)}}{tx}dx~dt$

$=\gamma e^{bc}E_1(b(a+c))+\int_1^a\dfrac{e^{bc}E_1(b(a+c))}{t}dt+\int_0^a\dfrac{(e^t-1)e^{bc}E_1(b(a+c))}{t}dt+\int_a^\infty\dfrac{e^{bc}e^tE_1(b(t+c))}{t}dt$

$=\gamma e^{bc}E_1(b(a+c))+e^{bc}E_1(b(a+c))\ln a+e^{bc}E_1(b(a+c))\int_0^a\dfrac{e^t-1}{t}dt+\int_a^\infty e^{bc}e^tE_1(b(t+c))~d(\ln t)$

$=e^{bc}E_1(b(a+c))\text{Ei}(a)+\int_{\ln a}^\infty e^{bc}e^{e^t}E_1(b(e^t+c))~dt$

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