I wondered what would happen if I integrated a function defined in terms of it’s derivative and after some puzzling, this is what I got:

Define fn(x)=−x⋅f′n−1(x)f_n(x)=-x\cdot f_{n-1}'(x). Now:

I=∫fn(x)dxI=\int f_n(x)\,dx

We take u=fn(x)⟹du=f′n(x)dxu=f_n(x)\implies\,du=f_n'(x)\,dx and dv=d⟹v=x\,dv=\,d\implies v=x, so:

I=x⋅fn(x)−∫x⋅f′n(x)dx=x⋅fn(x)+∫fn+1(x)dxI=x\cdot f_n(x)-\int x\cdot f_n'(x)\,dx=x\cdot f_n(x)+\int f_{n+1}(x)\,dx

which would mean:

∫fn(x)dx=x∞∑k=nfk(x)+C\int f_n(x)\,dx=x\sum_{k=n}^{\infty}f_k(x)+C

However, the above result doesn’t seem correct. Take for instance fn(x)=(n−1)!(logx)−nf_n(x)=(n-1)!(\log x)^{-n}.

It satisfies fn(x)=−x⋅f′n−1(x)f_n(x)=-x\cdot f’_{n-1}(x), but when I use try my little trick, I get:

∫(logx)−1dx=∫f1(x)dx=x∞∑k=1(k−1)!(logx)k+C\int(\log x)^{-1}\,dx=\int f_1(x)\,dx=x\sum_{k=1}^{\infty}\dfrac{(k-1)!}{(\log x)^k}+C

which isn’t correct.

Question: What did I do wrong? I really have no idea. I don’t use integration often, so it could be something realy obvious (I hope not).

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1

You forgot the limn→∞∫fn+1dx\lim_{n\to \infty}\int f_{n+1}\,dx term, “so to say”. But besides that the argument is indeed flawless for finite sums. I.e you indeed get: ∫fndx=xN∑k=nfk+∫fN+1dx+C, for any N∈N\int f_n\, dx=x\sum_{k=n}^N f_k +\int f_{N+1}\, dx+C, \text{ for any }N\in\mathbb{N}

– b00n heT

2 days ago

1

You didn’t take convergence into account. If the series ∑∞k=nfk(x)\sum_{k = n}^{\infty} f_k(x) converges nicely, then it works. If it doesn’t, you can only ever do finitely many such steps.

– Daniel Fischer♦

2 days ago

Okay, thanks. I now know what was wrong.

– Mastrem

2 days ago

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