I'm currently studying Calculus from the book "Introduction to Calculus and Analysis" by Richard Courant and Fritz John.
I started reading the chapter on Integration, and I stumbled upon the following problem.
After defining the integral as the limit of a sum, the authors derive the formula for the integral $\int_a^b x^\alpha$ (where $0<a<b$ and $\alpha$ is any positive integer) using geometric substitution. That is, instead of using sums based on equidistant points of division, the authors proceed by subdividing the interval $[a,b]$ by the points of a geometric progression: $$a,\ aq, \ aq^2, \ \ldots \ , \ aq^{n-1},\ aq^n = b.$$
Before this, they derive the formula for the integrals $\int_a^b x$ and $\int_a^b x^2$ using arithmetic substitution; that is, by considering equidistant points of division: $$a,\ a+h, \ a+2h, \ \ldots \ , \ a+(n-1)h, \ a+nh = b.$$
The authors suggest that this same argument could be used to derive the formula for the general case $\int_a^b x^\alpha$. However, since geometric substitution furnishes a simpler way of calculating the integral, they don't choose this method.
Therefore, in order to practice, I tried to compute the integral using arithmetic substitution:
- First, I wrote down the expression for the sum of $n$ rectangular areas, dividing the interval $[a,b]$ into $n$ equal parts: $$F_n = \sum_{i=1}^n (a+ih)^\alpha h$$
- Secondly, I tried to group terms: $$ F_n = \sum_{k=0}^\alpha \left[ {\alpha \choose k} \ a^{\alpha - k} \ h^{k+1} \ \left(\sum_{i=1}^n i^k \right) \right]$$
- Then, by letting $h = \frac{b-a}{n}$ and passing to the limit as $n \to \infty$, we have: $$ \lim\limits_{n \to \infty} F_n = \sum_{k=0}^\alpha \left[{\alpha \choose k} \ a^{\alpha - k} \ (b-a)^{k+1} \ \lim\limits_{n \to \infty} \left(\frac{1}{n^{k+1}} \sum_{i=1}^n i^k \right)\right]$$
- That inner limit is equal to $\frac{1}{k+1}$. Hence, $$\lim\limits_{n \to \infty} F_n = \sum_{k=0}^\alpha \left[{\alpha \choose k} \ a^{\alpha - k} \ (b-a)^{k+1} \frac{1}{k+1}\right],$$ which, in turn, is supposedly equal to $\frac{1}{\alpha + 1} (b^{\alpha + 1} - a^{\alpha + 1})$ (the formula that Courant and John got using geometric substitution), if I haven't done any mistake.
Unfortunately, despite the fact that the right-hand side of my last equality has the same pattern of Courant-John previous calculations when they computed $\int_a^b x$ and $\int_a^b x^2$, I don't see an easy way to assert that it equals $\frac{1}{\alpha + 1} (b^{\alpha + 1} - a^{\alpha + 1})$.
Could you please help me? First of all, is my reasoning above correct?
Edit 1: While reviewing my question, I noticed that I forgot to mention that for each interval of division $i$, I chose its right-hand end point as the intermediate value $\xi_i$ for computing the height of the rectangle. That is, if $f(x) = x^\alpha$, then for each interval $i$, $f(\xi_i) = (a+ih)^\alpha$ is the height of the corresponding rectangle, and its corresponding base is $h = \frac{b-a}{n}.$
Edit 2: After some thinking, I feel that one could prove that $$\sum_{k=0}^\alpha \left[{\alpha \choose k} \ a^{\alpha - k} \ (b-a)^{k+1} \frac{1}{k+1}\right] = \frac{1}{\alpha + 1} (b^{\alpha + 1} - a^{\alpha + 1})$$ by induction with respect to $\alpha$ and using some identity relating binomial coefficients like ${\alpha + 1 \choose k} = {\alpha \choose k-1} + {\alpha \choose k}$, but I'm having trouble in the induction step. The base case is straight-forward and I even computed for $\alpha = 2$, and it matches the formula for the integral $\int_a^b x^2$. If that's the case, could someone help with the induction step? Also, is there a more direct way by observing some property on the sum?
I'm adding the tag "induction" to my question.