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.

  • $\begingroup$ 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

$$\int e^{1/x} \, dx=e^{1/x} x-\text{Ei}\left(\frac{1}{x}\right) + c$$

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.

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

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 $x = 1/u$: $$\displaystyle I = xe^{1/x} - \int{\dfrac{ue^{u}}{u^2} \ du} = xe^{1/x} - \int{\dfrac{e^{u}}{u} \ du}$$ $$\boxed{\displaystyle I = xe^{1/x} - \operatorname{Ei} \left( \dfrac{1}{x} \right) + C}$$


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.

  • $\begingroup$ Okay. I can solve it then. Thanks:) $\endgroup$ – MB17 Dec 1 '17 at 12:51

There is no definite formula for this you can use series expansion of $e^x$ to compute a numerical value.

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

  • $\begingroup$ 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. $\endgroup$ – Sonal_sqrt Dec 1 '17 at 6:18
  • $\begingroup$ Okay. I got it. Thank you:) $\endgroup$ – MB17 Dec 1 '17 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.