Perplexing integral

Question

First and foremost, is it possible to get the integral you are trying to solve as the solution? I just got the same integral twice. I have also tried MATLAB but it gives the same result. Below is the integral i was solving:

$$\int_0^1e^{-xt} \left(e^{-t} + \frac{e^{-(t+1)} - 1} {1+t}\right) dt $$

It came about as a result of trying to solve the following fredholm equation by using the adomian decomposition method.

$$ u(x) = e^{-x} + \frac{e^{-(x+1)} - 1}{1+x} + \int_0^1e^{-xt}u(t)dt $$

Thanks a lot.

Answer 1

$$\int_0^1 \frac{e^{-xt}}{t + 1} dt = e^x\left(\Gamma(0,x)-\Gamma(0,2x)\right)$$

Where $$\Gamma(a, x) = \int_x^{\infty} t^{a - 1}e^{-t}dt$$ is the incomplete gamma function.

Answer 2

$$\int_0^1e^{-xt} \left(e^{-t} + \frac{e^{-(t+1)} - 1} {1+t}\right) dt =I_1+I_2+I_3$$ $I_1=\int_0^1e^{-(x+1)t} dt =\frac{1-e^{-(x+1)}}{x+1}$

$I_2=\int_0^1\frac{e^{-(xt+t+1)}}{1+t} dt =e^x\left(\Gamma(0,x)-\Gamma(0,2x) \right)$ with the incomplete Gamma function.

$I_3=\int_0^1 \frac{ - e^{-xt}} {1+t} dt =-e^x\left(\Gamma(0,x+1)-\Gamma(0,2x+2) \right)$

$\int_0^1e^{-xt} \left(e^{-t} + \frac{e^{-(t+1)} - 1} {1+t}\right) dt =\frac{1-e^{-(x+1)}}{x+1} + e^x\left(\Gamma(0,x)-\Gamma(0,2x)-\Gamma(0,x+1)+\Gamma(0,2x+2) \right)$