Ratio between two integrals: $$\frac {\int f(x)g(x)} { \int f(x)}.$$
Does exist a rule or do you know a way to solve it?
$f(x)= (1+x)^n e^{-ax}$
$g(x)= \ln(1+x)$
So: the numerator is the integral of $f(x)g(x)$ and the denominator is only the integral of $f(x)$.
And the question is if there exists a solution of this ratio..
My complete problem is: $$\frac {\int f(x)g(x)} {\left( \int f(x)\right)^2}.$$
On
The first integral $$\int{\ln(1+x)\left(1+x\right)^n e^{-ax}dx} $$ can be solved in the following way: $$\int{\ln(1+x)\left(1+x\right)^n e^{-ax}dx}=\left\{u=\ln(1+x);dv=(1+x)^{n}e^{-ax}dx\right\}=\ln(1+x)A(x)-\int\frac{1}{1+x}A(x)dx $$ Here $$A(x)=\int\left(1+x\right)^n e^{-ax}dx $$ $A(x)$ can be found by n times integrating by parts $$A(x)=\int\left(1+x\right)^n e^{-ax}dx =\frac{-1}{a}(1+x)^n e^{-ax}-\int n(1+x)^{n-1}e^{-ax}dx$$ The final answer is: $$A(x)=\sum\limits_{k=0}^{n}\frac{(-1)^{k+1}}{a^{k+1}}\frac{n!}{(n-k)!}(1+x)^{n-k}e^{-ax} $$ Moreover,$$\int f(x)dx=A(x)$$ To find $$\int \frac{A(x)}{1+x}dx $$ you can repeat previous integration all times except only $$ \int \frac{e^{-ax}}{1+x}dx$$ With the help of substitution it can be changed to something like $$\int\frac{e^{-bx}}{x}dx .$$ But it is not evaluated in elementary functions.
On
I can not calculate the Integral $$ \int f(x) g(x) \, dx $$ But, $$ \int f(x) \, dx=-e^{\alpha } (x+1)^{n+1} (\alpha (x+1))^{-n-1} \Gamma (n+1,(x+1) \alpha ) $$ where $\Gamma(a,z)$ is incomplete gamma function.
This may be helpful for you.
In the case of definite integrals, we have the following theorem, called the weighted mean value theorem: If $f$ and $g$ are continuous on a closed interval $[a,b]$, and if $g$ never changes sign in $[a,b]$, then we have $$ \int_{a}^{b} f(x) \cdot g(x) \ dx = f(c) \cdot \int_{a}^{b} g(x) \ dx $$ for some $c$ in $[a,b]$. For reference, see Calculus vol. 1 by Tom M. Apostol, Theorem 3.16. Hope this piece of information is of help.