Convergence and Conditions for Integrating an Infinite Series Solution

[bassota]

hey. i got an integral:
INT = integral

INT (e^-x)/x dx

PLEASE HELP

 

[ice109]

make a substitution

 

[bassota]

sorry, but i do not see how will a substitution help. i ahve tried it. i have also tried the integration by part and did not get anywhere. please explain, and show some details. thanks

 

[sutupidmath]

i do not think this has any nice solution, i mean in terms of any elementary function. I believe it does not have any closed form, unless we want to expand the function under integral sign using power series, and then integrate term per term, we would get an approximation of it. Use this link http://mathworld.wolfram.com/ExponentialIntegral.html

 

[dhris]

Doesn't look like that has a representation in terms of elementary functions. Check out the exponential integral function:
http://mathworld.wolfram.com/ExponentialIntegral.html

 

[DanielleL5432]

Can't you just use integration by parts over and over again...let u=1/x, then 1/x^2, etc and keep letting dv=e^-x dx?

 

[dhris]

i do not think this has any nice solution, i mean in terms of any elementary function. I believe it does not have any closed form, unless we want to expand the function under integral sign using power series, and then integrate term per term, we would get an approximation of it. Use this link http://mathworld.wolfram.com/ExponentialIntegral.html

Ha, you beat me to it. Can you believe it took me at least 6 minutes to write my two-line post?

Can't you just use integration by parts over and over again...let u=1/x, then 1/x^2, etc and keep letting dv=e^-x dx?

Sure, but what does that get you?

 

[HallsofIvy]

Can't you just use integration by parts over and over again...let u=1/x, then 1/x^2, etc and keep letting dv=e^-x dx?

Yes, that would give you an infinite series solution.

I think it would be easier to start with an infinite series:
$$e^{-x}= 1- x+ \frac{1}{2}x^2- \cdot\cdot\cdot\+ \frac{(-1)^n}{n!}x^n+\cdot\cdot\codt$$
so
$$\frac{e^{-x}}{x}= \frac{1}{x}- 1+ \frac{1}{2}x- \cdot\cdot\cdot\+ \frac{(-1)^n}{n!}x^{n-1}+\cdot\cdot\codt$$

Integrate that to get an infinite series solution.

 

[sutupidmath]

Yes, that would give you an infinite series solution.

I think it would be easier to start with an infinite series:
$$e^{-x}= 1- x+ \frac{1}{2}x^2- \cdot\cdot\cdot\+ \frac{(-1)^n}{n!}x^n+\cdot\cdot\codt$$
so
$$\frac{e^{-x}}{x}= \frac{1}{x}- 1+ \frac{1}{2}x- \cdot\cdot\cdot\+ \frac{(-1)^n}{n!}x^{n-1}+\cdot\cdot\codt$$

Integrate that to get an infinite series solution.

Wouldn't he need firs to determine whether this series converges uniformly, and whether they fullfill some other conditions which i do not actually know exactly which are they? Or it is obvious that this series can be integrated term by term?

P.S. Pardone my ignorance!