how to evaluate integral with different powers I am trying to evaluate the following
\begin{equation}
I(a,b) = \int_{a}^{\frac{a+b}{2}} (x-a)^{\alpha-1} \, x^n  \, dx + \int_{\frac{a+b}{2}}^{b} (b-x)^{\alpha-1} \, x^n \, dx,
\end{equation}
where $0<\alpha<1$. Wolfram alpha gives no solution. I tried integration by parts without success. My problem is that I don't understand well the evaluation of the limit of the upper limit and this integrand.
 A: Since the wording of the question was modified, my first anser is no longer valid. So, I post a new answer to the new wording :
\begin{equation}
I(a,b) = \int_{a}^{\frac{a+b}{2}} (x-a)^{p-1} \, x^n  \, dx + \int_{\frac{a+b}{2}}^{b} (b-x)^{p-1} \, x^n \, dx,
\end{equation}
About the convergence of the first integral :
Since $\quad p-1>-1\quad$ the integral is convergent at the lower bound : 
$$\int_{a}^{X\to\: a} (x-a)^{p-1} \, x^n  \, dx \sim a^n\frac{(X-a)^p}{p}$$
This is easy to prove with change of variable $\quad x=a+\epsilon$
Obviously, there is no problem of convergence at the upper bound insofar $b>a$. So, there is no problem of convergence for the first integral.
About the convergence of the second integral :
Since $\quad p-1>-1\quad$ the second integral is convergent at the upper bound : 
$$\int_{X\to \:b}^{b} (b-x)^{p-1} \, x^n \, dx\sim b^n\frac{(b-X)^p}{p}$$
This is easy to prove with change of variable $\quad x=b-\epsilon$
Obviously, there is no problem of convergence at the lower bound. So, there is no problem of convergence for the second integral.
NOTE :
These integrals cannot be expressed with a finite number of elementary functions. Some possible ways of solving are :


*

*Numerical calculus (suggested for technical applications).

*Solving in terms of infinite series for theory and limited series in practice.

*Solving in terms of special functions : The Beta and Incomplete Beta functions.
$$I(a,b)=a^{n+p}\left(\text{B}_{\frac{a+b}{2a}}(n+1,p)- \text{B}(n+1,p)\right) + b^{n+p}\left( \text{B}(n+1,p)-\text{B}_{\frac{a+b}{2b}}(n+1,p)\right) $$
A: Why did you separate the integration into two pieces? If I did not read anything wrong we state the integral as 
$$
I(a,b) = \int_{a}^{b} (x-a)^{\alpha-1}x^ndx
$$
Please correct me if this is wrong. Then, by change of variables (it may not be necessary but for the ease of calculation) and assuming that $n$ is integer, we can write the above integral as
$$
I(a,b) = \int_{0}^{b-a} t^{\alpha-1}(t+a)^ndt
$$
where I used $t=x-a$. By using the binomial theorem $(t+a)^n=\sum_{k=0}^n\binom{n}{k}t^ka^{n-k}$, we can write $I(a,b)$ as
$$
I(a,b) = \int_{0}^{b-a} t^{\alpha-1}\sum_{k=0}^n\binom{n}{k}t^ka^{n-k}dt\implies \sum_{k=0}^n\binom{n}{k}a^{n-k}\int_{0}^{b-a}t^{k+\alpha-1}dt \\
\implies I(a,b)= \sum_{k=0}^n\binom{n}{k}a^{n-k}\frac{t^{k+\alpha}}{k+\alpha}|^{b-a}_0= \sum_{k=0}^n\binom{n}{k}a^{n-k}\frac{(b-a)^{k+\alpha}}{k+\alpha}
$$
If $n$ is a real number then the summation becomes infinite as stated at http://mathworld.wolfram.com/BinomialTheorem.html. Let me know if there are any errors in my derivation.
A: \begin{equation}
I(a,b) = \int_{a}^{\frac{a+b}{2}} (x-a)^{p-1} \, x^n  \, dx + \int_{\frac{a+b}{2}}^{b} (x-a)^{p-1} \, x^n \, dx \qquad 0<p<1.
\end{equation}
I wonder if where is a typo in the equation : Why you don't simply wrote
\begin{equation}
I(a,b) = \int_{a}^{b} (x-a)^{p-1} \, x^n  \, dx  \qquad 0<p<1.
\end{equation}
Supposing that $\quad b>a\quad$ so that $\quad (x-a)^{p-1}\quad$ be real on the range $\quad a<x<b$
Since $\quad p-1>-1\quad$ the integral is convergent at the lower bound : 
$$I(a,b\to a)\sim a^n\frac{(b-a)^p}{p}$$
This is easy to prove with change of variable $\quad x=a+\epsilon$
Moreover, I cannot see any problem of convergence at the upper bound insofar $b>a$.
For further calculus, HINT : The integral involves the Incomplete Beta function.
ATTENTION :
As suspected, there was a mistake in the initial wording of the question.
The typo was corrected by Math. So, My answer to the initial question is no longer consistent with the new wording.
Anyways, the HINT remains valid : The integrals involve the Incomplete Beta function.  
