First, thanks for your time and input. In some research, I evaluate the following integral $$\int_0^1 x (n-1) n v^{-2n}(v-x)\lambda^{-n}(v-2x+\lambda v)(\lambda v-x)(x(v+\lambda v-x))^{n-2} dx$$ where v>0 $\lambda$>1 and n>2. using Mathematica and the solution is very ugly (including hypergeometric functions).
I was wondering if anyone could help advise me on how to start to evaluate this integral by hand.
Thanks!
Well, simply rewrite your function. Here's how to do it by hand (as you asked):
First let's write:
$$\displaystyle I=\int_0^1 x (n-1) n v^{-2n}(v-x)\lambda^{-n}(v-2x+\lambda v)(\lambda v-x)(x(v+\lambda v-x))^{n-2} dx=\int_0^1f(x)dx$$
I'll just get rid of all the constant terms and put them on the side:
$f(x)=\left((n-1)nv^{-2n}\lambda^{-n}\right)\left(x(v-x)(v-2x+\lambda v)(\lambda v-x)x^{n-2}(v+\lambda v-x)^{n-2}\right)$
Now let's introduce some notations:
$$\left\{\begin{array}{cc}\left((n-1)nv^{-2n}\lambda^{-n}\right)=A \\ v+\lambda v=B\end{array}\right.$$
Replace in $I$:
$$\displaystyle I=\int_0^1 Ax(v-x)(B-2x)(\lambda v-x)x^{n-2}(B-x)^{n-2}dx$$
Now get rid of this constant term $A$ and call this new function in the integral $g$:
$$\displaystyle I=A\int_0^1 (v-x)(B-2x)(\lambda v-x)x^{n-1}(B-x)^{n-2}dx=A\int_0^1g(x)dx$$
Expand $g$:
$$g(x)=(v-x)(B-2x)(\lambda v-x)x^{n-1}(B-x)^{n-2}$$ $$=(vB-(B+2v)x+2x^2)(\lambda v-x)x^{n-1}(B-x)^{n-2}$$ $$=((v^2B\lambda)-x(vB+(B+2v)\lambda v)+x^2(2\lambda v+B+2v)-2x^3)x^{n-1}(B-x)^{n-2}$$
Add some more notations (just because I don't want to carry all these constants around):
$$\left\{\begin{array}{cc}(v^2B\lambda)=C_1 \\ (vB+(B+2v)\lambda v)=C_2 \\ (2\lambda v+B+2v)=C_3 \end{array}\right.$$
Your function looks a lot better now:
$$g(x)=\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)(B-x)^{n-2}$$
The "hard" part is the $(B-x)^{n-2}$, but not that hard actually, because you can write it this way:
$$\displaystyle(B-x)^{n-2}=\sum_{k=0}^{n-2}{n-2 \choose k}B^{n-2-k}(-x)^k=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}x^k$$
So $g$ becomes:
$$g(x)=\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}x^k$$
$$g(x)=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{n-1}-C_2x^{n}+C_3x^{n+1}-2x^{n+2}\right)x^k$$
$$g(x)=\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right)$$
Let's go back to the integral:
$$\displaystyle I=A\int_0^1 g(x)dx$$ $$\displaystyle I=A\int_0^1 \sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$$ $$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\int_0^1 \left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$$ $$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\times J_k$$
where $\displaystyle J_k=\int_0^1 \left(C_1x^{k+n-1}-C_2x^{k+n}+C_3x^{k+n+1}-2x^{k+n+2}\right) dx$
Now you have a simple polynomial to integrate:
$$\displaystyle J_k=\left[\frac{C_1}{k+n}x^{k+n}-\frac{C_2}{k+n+1}x^{k+n+1}+\frac{C_3}{k+n+2}x^{k+n+2}-\frac{2}{k+n+3}x^{k+n+3}\right]_0^1$$
$$\displaystyle J_k=\frac{C_1}{k+n}-\frac{C_2}{k+n+1}+\frac{C_3}{k+n+2}-\frac{2}{k+n+3}$$
And thus you have:
$$\displaystyle I=A\sum_{k=0}^{n-2}(-1)^k{n-2 \choose k}B^{n-2-k}\left(\frac{C_1}{k+n}-\frac{C_2}{k+n+1}+\frac{C_3}{k+n+2}-\frac{2}{k+n+3}\right)$$
You can replace your values $A$,$B$,$C_1$,$C_2$ and $C_3$ and get your result (which I won't write here).
I admit it's still pretty ugly but at least you only have simple fractions and sums (and binomial terms). But with a function like yours you can't just expect a nice and clean solution ;)