For a given real number \(\alpha\) and \(n, n\ge 0\). There is a trick to evaluate:
\[\begin{align} I_n &= \int_0^1{\frac{x^n}{x+\alpha} dx}\\ &= \int_0^1 \frac{x^n + x^{n-1}\alpha - x^{n-1} \alpha}{x+\alpha} dx \\ &= \int_0^1 \frac{x^n + x^{n-1}\alpha}{x+\alpha} dx - \int_0^1 \frac{x^{n-1}\alpha}{x+\alpha} dx\\ &= \int_0^1 \frac{x^{n-1}(x+\alpha)}{x+\alpha} dx - \int_0^1 \frac{x^{n-1}\alpha}{x+\alpha} dx\\ &= \int_0^1 x^{n-1} dx - \alpha I_{n-1}\\ &= \frac{1}{n} x^n \bigg|^1_0 - \alpha I_{n-1}\\ &= \frac{1}{n} - \alpha I_{n-1}\\ I_n &= \frac{1}{n} - \alpha I_{n-1} \end{align}\]
So, to evaluate \(I_{100}\):
\[\begin{align} I_1 &= \frac{1}{1} - \alpha I_0\\ I_2 &= \frac{1}{2} - \alpha I_1\\ ...\\ I_{100} &= \frac{1}{100} - \alpha I_{99}\\ \end{align}\]
\[\begin{align} I_0 &= \int_0^1 \frac{x^0}{x+\alpha} dx\\ &= \int_0^1 \frac{1}{x+\alpha} dx\\ &= \ln(x+\alpha) \bigg|^{x=1}_{x=0}\\ &= \ln(1+\alpha)-\ln(\alpha)\\ \end{align}\]
In matlab:
% Try alpha values of 0.5 and 2.
alpha = 0.5;
N = 100;
I = log((1+alpha) / alpha);
for n = 1:N
I = 1/n - alpha * I;
end
disp(['Answer: ' num2str(I)]);If \(\alpha=0.5\), \(I_{100}=0.0066444\).
If \(\alpha=2\), \(I_{100}=6.053*10^{12}\).
For \(0 \le x \le 1, \alpha \gt 1\):
\[0 \le \frac{x^n}{x+\alpha} \le x^n \Rightarrow \int_0^1 \frac{x^n}{x+\alpha} \le \int_0^1 x^n\]
This means \(I_{100}\) should have been less than \(\frac{1}{101}\), and not on the order of \(10^{12}\).
Each computed \(I\) value has an error associated with it that compounds over multiple operations.
\[\begin{align} I_n^{comp} &= I_n^{exact} + e_n\\ I_n^{exact} &= \frac{1}{n} - \alpha I^{exact}_{n-1}\\ I_n^{comp} &= \frac{1}{n} - \alpha I^{comp}_{n-1}\\ e_n &= I_n^{comp} - I_n^{exact}\\ &= -\alpha(I_{n-1}^{comp} - I_{n-1}^{exact})\\ &= -\alpha e_{n-1}\\ &= -\alpha(-\alpha e_{n-2})\\ |e_n| &= |\alpha| |e_{n-1}|\\ &= |\alpha|^2 |e_{n-2}|\\ &= |\alpha|^3 |e_{n-3}|\\ ...\\ &=|\alpha|^n |e_0|\\ \end{align}\]
If \(|\alpha|<1\), then \(|e_n| \rightarrow 0\) (good)
If \(|\alpha|>1\), then \(|e_n| \rightarrow \infty\) (bad)