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There is a theorem that the power series can be integrated term by term over any closed and bounded interval contained in the interval of convergence.

I assume I am using this theorem when I obtain the series expansion of $ \tan^{-1} (x)$ by integrating each term in the power series expansion of $ \frac{1}{1+x^2} $. But this theorem fails me when I apply it to the power series expansion of $ \ e^x $

Since $$ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}.....(1) $$ Now integrating both sides $$ \int e^xdx=\int 1\ dx+\int x\ dx+\int \frac{x^2}{2!}\ dx....$$ $$ e^x=x+\frac{x^2}{2!}+\frac{x^3}{3!}....(2) $$ I belive eq (1) and eq(2) are different. I know that I haven't added the constant of integration. That is because as I mentioned earlier the series expansion of $ tan^{-1}x $ i obtained by integration of series expansion of $ \frac{1}{1+x^2} $ and no integration constant was added there. And I know that expansion is correct because I got that from "tom apostol calculus 1" $$ \int\frac{1}{1+x^2}=\int1-\int x^2+\int x^4-\int x^6.... (3)$$ $$ tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}......(4) $$ So how is eq (4) supposed to be right if eq(2) is wrong.


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  • $\begingroup$ Integration constant? $\endgroup$ – Peter Szilas Jan 29 '20 at 8:43
  • $\begingroup$ Where is the constant? $\endgroup$ – Fakemistake Jan 29 '20 at 8:44
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$$\int_0^x e^tdt=e^x\color{red}{-1}$$ matches your term-wise integration. (While $\displaystyle\int_0^x t^ndt=\dfrac{x^{n+1}}{(n+1)!}$, constant term is $0$.)

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  • $\begingroup$ the series expansion of tan inverse is obtained by a similar integration of an expansion which does not involve adding a constant. I have explained it in detail in an edit. Could you please read it and explain why the first one is wrong but the second one right $\endgroup$ – Siddharth Prakash Jan 29 '20 at 14:28
  • $\begingroup$ @SiddharthPrakash: if you understood my answer, you know that the constant is $\arctan(0)$. $\endgroup$ – Yves Daoust Jan 29 '20 at 14:29
  • $\begingroup$ I get your answer now thank you very much $\endgroup$ – Siddharth Prakash Jan 29 '20 at 14:38
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You have to add a constant when you take anti-derivative. The two expression you got for $e^{x}$ are some except for the constant term so there is no contradiction.

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  • $\begingroup$ the series expansion of tan inverse is obtained by a similar integration of an expansion which does not involve adding a constant. I have explained it in detail in an edit. Could you please read it and explain why the first one is wrong but the second one right $\endgroup$ – Siddharth Prakash Jan 29 '20 at 14:27
  • $\begingroup$ @SiddharthPrakash In the second case you got a correct result by a wrong method! The constant of integration can happen to be $0$ in some cases but it is wrong to omit the constant. $\endgroup$ – Kavi Rama Murthy Jan 29 '20 at 23:23
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You're forgetting the +C!

Though if you want a better way of going about this, just try using definite integrals.

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  • $\begingroup$ the series expansion of tan inverse is obtained by a similar integration of an expansion which does not involve adding a constant. I have explained it in detail in an edit. Could you please read it and explain why the first one is wrong but the second one right $\endgroup$ – Siddharth Prakash Jan 29 '20 at 14:27
  • $\begingroup$ As for the inverse tangent expansion you found, it does involve adding a constant, it just happens that $C=0$. You have just found the right answer by accident in this case. $\endgroup$ – Isaac Browne Jan 29 '20 at 16:59

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