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.

  • 2
    $\begingroup$ It is quite possible that a solution technique will fail, just giving you the same problem over again. But can you explain what you mean by "solving" the integral? This integral can be expressed in terms of the incomplete gamma function. $\endgroup$ – Paul Sinclair Dec 28 '15 at 20:30
  • $\begingroup$ what exactly is that? Because as far as i know for now, i have to iteratively compute the above integral while plugging in the results for from the previous caculation to obtain a new solution. besides, have no idea why even MATLAB is giving me the same solution all along. its also giving exactly the same integral presented above as the solution to that same integral. is that normal? $\endgroup$ – guthik Dec 28 '15 at 21:12
  • $\begingroup$ MATLAB is a program. Programs are stupid. They blindly follow a recipe given them, and if it doesn't work, then there is nothing more they can do. In this case, The algorithm MATLAB is following doesn't work. You, however, are not stupid. When you see it doesn't work, you should abandon it and look for a different approach that might work. $\endgroup$ – Paul Sinclair Dec 28 '15 at 21:24

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

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

  • $\begingroup$ Wow, honestly i have solved quite a number of integrals till this point but this is the first time am seeing something like this :) $\endgroup$ – guthik Dec 28 '15 at 21:33
  • $\begingroup$ The resulting two integrals are indeed way very easy to solve, but i don't see how to connect the results inorder to get the final answer to the original problem. $\endgroup$ – guthik Dec 28 '15 at 21:36
  • $\begingroup$ Sinclari. Thanks a lot. you guys are the best :). Now i can take it from here. But i dont know which answer to accept. they are both very helpful. $\endgroup$ – guthik Dec 29 '15 at 10:32
  • $\begingroup$ I should admit that I got that particular expression from Wolfram Alpha. These sorts of things are exactly why reference works were invented. Of course, I only looked it because I broke your integral down as per JJacquelin's post. $\endgroup$ – Paul Sinclair Dec 29 '15 at 14:05

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

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

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

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

  • $\begingroup$ Thanks a lot. you guys are the best :). Now i can take it from here. But i dont know which answer to accept. they are both very helpful. $\endgroup$ – guthik Dec 29 '15 at 10:31

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