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Iam trying to solve this as a part of some problem. But I am not able to find the value for this.

Can anyone integrate the above problem.

Please let me know. Thank you.

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  • $\begingroup$ Do you have any suggestion? What have you done yourself? $\endgroup$ – Cardinal Dec 1 '17 at 5:58
  • $\begingroup$ I tried by various substitutions but I just arrive at some dead end. $\endgroup$ – MB17 Dec 1 '17 at 6:02
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$$\int e^{1/x} \, dx=e^{1/x} x-\text{Ei}\left(\frac{1}{x}\right) + c$$

where $\operatorname{Ei}(x)$ is the exponential integral function defined as

$$\operatorname{Ei}(x) = - \int_{-x}^\infty \frac{e^{-t} dt}t$$

$\operatorname{Ei}(x)$ is a non-elementary function, and we can only evaluate it using numerical methods.

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  • $\begingroup$ Sorry. I am not following it. You mean the solution doesnt exist? $\endgroup$ – MB17 Dec 1 '17 at 6:03
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    $\begingroup$ @MB17 No, just that the integral can't be expressed in terms of elementary functions $\endgroup$ – user223391 Dec 1 '17 at 6:04
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Make a substitution of $x = \tan \theta$, then you obtain

$$ I = \int e^{\cot \theta} \sec^2\theta \ \ d\theta $$

Integrate by parts with $u$ being the exponential:

$$ uv - \int{v \ du} = \tan \theta \ e^{\cot \theta} + \int{\dfrac{e^{\cot \theta}}{\sin \theta \cos \theta} \ d \theta} $$

Make a substitution of $u = \cot \theta \implies du = -\csc^2 \theta$, noting that $\tan \theta = 1/u$ (and substituting $x$ back in):

$$I = xe^{1/x} - \int{\dfrac{e^u}{u} \ du}$$

We arrive at the final answer by substituting everything back in:

$$\boxed{\displaystyle I = xe^{1/x} - \operatorname{Ei} \left( \dfrac{1}{x} \right) + C}$$

I explain more of the small algebraic details here, if any step is unclear.


A note on the solution: There's actually a much simpler solution I found after realizing the trig sub is redundant. Here it is:

Start by integrating by parts, with $dv = 1$: $$\displaystyle I = xe^{1/x} - \int{\dfrac{-xe^{1/x}}{x^2} \ dx} = xe^{1/x} + \int{ \dfrac{e^{1/x}}{x} \ dx}$$ We can substitute $u = 1/x \implies du = -1/x^2 \ dx$, noting that $x = 1/u$: $$\displaystyle I = xe^{1/x} - \int{\dfrac{ue^{u}}{u^2} \ du} = xe^{1/x} - \int{\dfrac{e^{u}}{u} \ du}$$ $$\boxed{\displaystyle I = xe^{1/x} - \operatorname{Ei} \left( \dfrac{1}{x} \right) + C}$$

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One simple way to do it (if an approximate solution will do) is the following:

You know the Taylor series of $e^x$ is given by $e^x = 1 + x + \frac{x^2}{2!}+...$.

Therefore the Taylor series of $e^{1/x}$ is given by $e^{1/x} = 1 + \frac{1}{x} + \frac{1}{2!x^2} +...$

Now just integrate term-by term.

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  • $\begingroup$ Okay. I can solve it then. Thanks:) $\endgroup$ – MB17 Dec 1 '17 at 12:51
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There is no definite formula for this you can use series expansion of $e^x$ to compute a numerical value.

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$$ $$\int e^{1/x}=\int1+\frac{1}{x}+\frac{1}{x^22!}+\frac{1}{x^33!}+\frac{1}{x^44!}+.... \\ \int e^x= C + x + \ln x-\frac{1}{2!}\frac{1}{x}-\frac{1}{3!}\frac{1}{2x^2}-\frac{1}{3!}\frac{1}{3x^3}-...$$

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  • $\begingroup$ So, There is no exact solution. Only approximated solution exists? $\endgroup$ – MB17 Dec 1 '17 at 6:07
  • $\begingroup$ @MB17 No there is an exact solution, it just can't be expressed in terms of elementary functions $\endgroup$ – user223391 Dec 1 '17 at 6:07
  • $\begingroup$ What is the exact solution then? In what way can it be expressed? $\endgroup$ – MB17 Dec 1 '17 at 6:10
  • $\begingroup$ The more terms in the series you sum the more accurate the answer becomes. $\endgroup$ – Piyush Divyanakar Dec 1 '17 at 6:18
  • $\begingroup$ Okay. I got it. Thank you:) $\endgroup$ – MB17 Dec 1 '17 at 12:49

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