Consider $\int e^x \mathrm{d}x$. Using integration by parts, and setting $u=e^x$, $v=1$, we find that $\int e^x \mathrm{d}x = xe^x-\int \frac 1 2 x^2 e^x \mathrm{d}x = xe^x - \frac 1 2 x^2 e^x + \int \frac 1 6 x^3 e^x \mathrm{d}x = \sum_{n=0}^{\infty} (-1)^n \frac {x^{n+1}e^x} {(n+1)!} + C$. Simplifying, we get $\int e^x \mathrm{d}x = e^x+C_1 = \sum_{n=0}^{\infty} (-1)^n \frac {x^{n+1}e^x} {(n+1)!} + C_2$. But $\sum_{n=0}^{\infty} (-1)^n \frac {x^{n+1}e^x} {(n+1)!} = e^x - 1$, which makes the statement above be correct iff $C_2 = C_1 + 1$. However, I thought that integration by parts always yielded statements where the constants were the same on both sides of the equation. Why is this occurring?
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9don't forget "$+C$"... You could also try your formal calculation as a definite integral: $\int_0^x e^t,dt$. – peter a g Oct 10 '23 at 00:32
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@peterag Make this an answer. – GEdgar Oct 10 '23 at 00:48
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Indefinite integrals are always computed up to an additive constant. – Abezhiko Oct 10 '23 at 04:17
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1This is addressed in this answer to a previous question. – JonathanZ Oct 10 '23 at 05:27
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As requested:
Don't forget the "+C" of indefinite integration. You could also try your formal calculation in the form of a definite integral: $$ \int^x_0 e^t\,dt.$$
peter a g
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