## anonymous one year ago Primitive !

1. anonymous

$\int\limits_{}^{}e^\frac{ -1 }{ t }$

2. anonymous

dt

3. anonymous

|dw:1436757696187:dw|

4. SolomonZelman

$$\large\color{black}{ \displaystyle \int e^{-1/t}~dt }$$ $$\large\color{black}{ \displaystyle u=-1/t }$$ $$\large\color{black}{ \displaystyle du=1/t^2~dt }$$ $$\large\color{black}{ \displaystyle t^2~du=dt }$$ $$\large\color{black}{ \displaystyle (-1/u)^2~du=dt }$$ $$\large\color{black}{ \displaystyle 1/u^2~du=dt }$$ $$\large\color{black}{ \displaystyle \int \frac{e^{u}}{u^2}~du }$$ I don't think this has an answer in term of a simple function. What I see to do instead is: $$\large\color{black}{ \displaystyle e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!} }$$ $$\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1/t)^n}{n!} }$$ $$\large\color{black}{ \displaystyle e^{-1/t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} }$$ $$\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\int \sum_{n=0}^{\infty}\frac{(-1)^nt^{-n}}{n!} dt }$$ $$\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^nt^{-n+1}}{~(-n+1)~n!~} +C }$$ and that you can simplify as: $$\large\color{black}{ \displaystyle \int e^{-1/t}~dt~=\sum_{n=0}^{\infty}\frac{(-1)^n}{~~t^{n-1}~(-n+1)~n!~} +C }$$

5. SolomonZelman

I mean if you haven't learned this technique using a series, then disregard my reply....

6. anonymous

I agree that there's no antiderivative in terms of elementary functions. This seems relevant: http://mathworld.wolfram.com/ExponentialIntegral.html You should be able to get that if you integrate by parts. (I haven't bothered checking for myself, though.)

7. SolomonZelman

Oh, I have seen a post with this idea of integration by parts a certain number of times after which you arrive at an infinite series. I have done a similar thing when integrated e^x/x. (of course differentiating the 1/x) This is how I got to know the series technique, but without proving/showing using by parts every time is more convinient:D