$\displaystyle \int \left(x \ln x - \frac{(n-1)!}{x^n} \right)e^x \ dx = ?$
I bet there is a better way , but I thought this was fun.
First lets break this up a bit
$\displaystyle
\int x\ln(x)e^xdx-(n-1)!\int x^{-n}e^xdx
$
Lets focus on the 2nd integral to start
$\displaystyle (n-1)!\int x^{-n}e^xdx$
so if we integrate by parts with $\displaystyle u=e^x \implies du =e^xdx$ and $\displaystyle dv=x^{-n} \implies v=-\frac{1}{n-1}x^{-n+1}$
$\displaystyle (n-1)!\int x^{-n}e^xdx=(n-1)!\left(-\frac{1}{n-1}x^{-n+1}e^x+\frac{1}{n-1}\int x^{-n+1}e^xdx \right)= $
$\displaystyle -(n-2)!x^{-n+1}e^{x}+(n-2)!\int x^{-n+1}e^x$
If we integrate by parts again we get
$\displaystyle -(n-2)!x^{-n+1}e^{x}-(n-3)!x^{-n+2}e^{x}+(n+3)!\int x^{-n+2}e^x$
Doing this n-1 times we get
$\displaystyle (n-1)!\int x^{-n}e^xdx=-e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+\int \frac{1}{x}e^{x}dx$
Now lets look at the last bit
$\displaystyle \int \frac{1}{x}e^{x}dx$ if we integrate this by parts we get
$\displaystyle \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-\int e^{x}\ln(x)dx$
Now lets integrate the last integral by parts again with
$\displaystyle u=e^{x} \implies du=e^{x},dv=\ln(x ) \implies v=x\ln(x)-x$
$\displaystyle \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-[x\ln(x)-x]e^{x}+\int [x\ln(x)-x]e^xdx$
$\displaystyle \int \frac{1}{x}e^{x}dx=e^{x}\ln(x)-[x\ln(x)-x]e^{x}+\int x\ln(x)e^{x}dx-\int xe^xdx $
So finally we get
$\displaystyle (n-1)!\int x^{-n}e^xdx=-e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)-[x\ln(x)-x]e^{x}+$
$\displaystyle \int x\ln(x)e^{x}dx-\int xe^xdx $
Now subbing this back into the very first line we get
$\displaystyle e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)+[x\ln(x)-x]e^{x}+\int xe^xdx $
evaluateing the last integral and simplifying gives
$\displaystyle e^{x}\sum_{i=1}^{n-1}[n-(1+i)]!x^{-n+i}+e^{x}\ln(x)+xe^{x}\ln(x)-e^x $
Hello,
Here is what I tried...
$\displaystyle I_n(x)=\int\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x ~dx$
By differentiating this (under the integral sign), we can get :
$\displaystyle I_n'(x)=I_n(x)+\int\left(1+\ln(x)+\frac{n!}{x^{n+1 }}\right)e^x~dx$
$\displaystyle I_n'(x)=I_n(x)-I_{n+1}(x)+\int\left(1+\ln(x)+x\ln(x)\right)e^x~dx$
But the latter is exactly $\displaystyle x\ln(x)e^x+C$
And we also have $\displaystyle I_n'(x)=\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x$
So finally :
$\displaystyle I_{n+1}(x)-I_n(x)=\frac{(n-1)!}{x^n}\cdot e^x+C$
And :
$\displaystyle I_n(x)=e^x \sum_{k=2}^n \frac{(k-2)!}{x^{k-1}}+I_2+D$
That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute $\displaystyle I_2$... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.
I'll think about it later. See ya
here Moo actually used by parts: $\displaystyle x \ln x - \frac{(n-1)!}{x^n} = u, \ e^x dx = dv$ and the fact that $\displaystyle uv=I_n'(x).$
this is a nice approach. you probably don't need to think about finding $\displaystyle I_2$ because $\displaystyle \emptyset$'s solution, which is exactly how i created the problem, has already answered your question.
$\displaystyle I_n'(x)=I_n(x)-I_{n+1}(x)+\int\left(1+\ln(x)+x\ln(x)\right)e^x~dx$
But the latter is exactly $\displaystyle x\ln(x)e^x+C$
And we also have $\displaystyle I_n'(x)=\left(x\ln(x)-\frac{(n-1)!}{x^n}\right)e^x$
So finally :
$\displaystyle I_{n+1}(x)-I_n(x)=\frac{(n-1)!}{x^n}\cdot e^x+C$
And :
$\displaystyle I_n(x)=e^x \sum_{k=2}^n \frac{(k-2)!}{x^{k-1}}+I_2+D$
That's where I've gotten to. I didn't have much time this morning. And since I'm not at home, I don't have a paper in front of me and cannot continue immediately... I think the problem will be to compute $\displaystyle I_2$... The sum is just the (n-1) (or (n-2)) th derivative of x/(x-1)=1+1/(x-1), which can easily be found.
I'll think about it later. See ya