It is well-known that we can represent a Harmonic number as the following integral:
$$H_n = \int_0^1 \frac{1-t^n}{1-t} dt$$
The derivation of this integral doesn't need you to derive the indefinite integral first, so now I'm wondering what the indefinite integral is and how one can derive it. According to WolframAlpha the indefinite integral is:
$$\int \frac{1-t^n}{1-t} dt = \frac{t^{n+1}{}_2F_1(1,n+1;n+2;t)}{n+1} – \ln(1-t) + C$$
where ${}_2F_1(a,b;c;z)$ is a Hypergeomtric function. I understand why $-\ln(1-t)$ is at the end, that's the result of splitting up the integrand, but I don't understand how a Hypergeometric function ends up there.
Best Answer
Note that for $n\in\mathbb{N}$, we have $$\frac{1-t^n}{1-t}=1+t+t^2+\cdots+t^{n-1}$$ It follows that $$\int\:\frac{1-t^n}{1-t}\:dt=t+\frac{t^2}{2}+\frac{t^3}{3}+\cdots+\frac{t^n}{n}+C$$