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As a follow up to a previous posting addressing the integral of $1/ (t^n+1)$ for $n\in \Bbb{N}$ I found the following

$$\int_0^ x \frac{1}{1+t^n}\, dt=\sum_{i=0}^{\infty}\frac{(i!)(n^i)x^{in+1}} {(x^n+1)^{i+1}\prod_{k=0}^i (kn+1)}$$

My son programmed the equation in CC++, and showed me that it works for $n=1,2$, 3... But as n gets large the computer cannot calculate.

  • Many thanks to the management of this webside.

I would like assistance verifying if this equation is correct or wrong.

My request is also for others to demonstrate how the equation may be developed.

The relevance of the equation is that it generates multiple series/functions. Think of it as a function generator n. E.g.

At n= 1 the equation gives the series for $$LN(x+1)$$

At n= 2 the equation gives the series for $$ATAN(x)$$

The equation should be useful to other.

Martin Sleziak
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926reals
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  • The sum is probably a taylor series expansion of the hypergeometric function. – 1110101001 Jun 24 '16 at 04:42
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    Integrating from 0 to $x$ with respect to $x$? :( –  Jun 24 '16 at 23:51
  • I corrected the equation – 926reals Jun 25 '16 at 00:15
  • I changed the $x$ to a $t$ for a reason. It doesn't make sense (or is at least confusing) to integrate wrt $x$ from $0$ to $x$. –  Jun 25 '16 at 00:26
  • if you change x to t on the left hand side then you must change x to t in the right side. Please leave the equation alone... The relationship between Ln(x+1) and Atan(x) is that from my equation at n=1 you get Ln(x+1) and at n=2 you get Atan(x). – 926reals Jun 25 '16 at 00:33
  • You cannot use the same variable in the integrand and limits of integration. Additionally power series when summed to infinity are NOT approximations, they are exact. –  Jun 25 '16 at 01:14
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    I changed the integration variable back to $t$. In the left hand side the "true" variable is the boundary of the integration, and, as you see, that is still $x$. The role of $t$ in the left is very similar to the role of $i$ on the right - they are both summation variables. One is discrete the other is continuous, but that is irrelevant. – Jyrki Lahtonen Jun 25 '16 at 05:58
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    Mind you, I've seen a chemistry professor make the same mistake in a textbook (using the integration variable as a boundary as well). You can guess what is meant (as we all did here), but it is just not kosher. And may lead to serious confusion, if you, say, differentiate the formula w.r.t. that variable. Compare: Is the intended meaning of $$\sum_{n=1}^nn$$ $1+2+3+\cdots+n=n(n+1)/2$ or $n+n+n+\cdots+n=n^2$? We cannot tell, and therefore the notation is deemed inappropriate. Same thing here. – Jyrki Lahtonen Jun 25 '16 at 06:09
  • This integral is particularly interesting as $x\to \infty$ - if you're interested in knowing what the result for that case is, you might have a look here: http://math.stackexchange.com/q/1814430/12864 – Christian Ivicevic Jun 30 '16 at 05:32

1 Answers1

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Where did you find that equation? It's quite different from what I got, which I shall explain now.

First, a common power series is

$$ \frac{1}{1-x} = \sum_{i\geq0} x^{i}. $$

Using the substitution $x=-t^{n}$,

$$ \frac{1}{1+t^{n}} = \sum_{i\geq0} (-t^{n})^{i} = \sum_{i\geq0} (-1)^{i}t^{in}. $$

Then,

$$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \int_{0}^{x} \sum_{i\geq0} (-1)^{i}t^{in} dt = \sum_{i\geq0} (-1)^{i} \int_{0}^{x} t^{in} dt. $$

The integral is simply

$$ \left[ \frac{t^{in+1}}{in+1} \right]_{0}^{x} = \frac{x^{in+1}}{in+1}. $$

Thus

$$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \sum_{i\geq0} (-1)^{i}\frac{x^{in+1}}{in+1}. $$

Any comments on where I may have gone wrong are appreciated.

Bilbottom
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  • @EstebanRamirezLopez When did they use an approximation? –  Jun 25 '16 at 00:26
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    @EstebanRamirezLopez Zachary Selk has made a valid point in the question comments - it is confusing to define the integral of some function of $x$ with respect to $x$ where $x$ is also a limit of the integral. – Bilbottom Jun 25 '16 at 00:32
  • @EstebanRamirezLopez Is your sum from $i=0$ to infinity? – Bilbottom Jun 25 '16 at 00:54
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    I would also like to point out that, when $n=2$, the integral is arctan($x$) using your notation. The power series of arctan($x$) is $\sum_{i\geq 0} (-1)^{i} \frac{x^{2i+1}}{2i+1}$ which is what you get when you substitute $n=2$ into my equation. I couldn't get to that using yours (maybe I did it wrong? It's the first time seeing that equation). Furthermore, power series only become approximations when they are truncated. An infinite sum is equal to the function that you are 'approximating'. – Bilbottom Jun 25 '16 at 01:02
  • Bilbottom You must have done something wrong in evaluating n= 2. Or you evaluated a previous equation I posted, which had a errors. (I'm knew to this site and having to learn the math-editor). Power series are; however, a coincidence of calculus, so I apologize, Power series are not an approximation, they are an example of a larger result. – 926reals Jun 25 '16 at 01:23
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    +1 this seems correct. But you should specify it only holds for $|x|<1$.. Also, why can one interchange the integral and the summation? – Ant Jun 25 '16 at 06:26
  • @Ant You are correct, I should have specified the radius of convergence, thank you for adding that. If you write the summation out you can see why the integral and summation can be swapped; let me demonstrate with an easier example: $$ \int \sum_{i\geq 0} x^{i} dx = \int 1 + x + x^{2} + ... dx $$ $$ \int 1 dx + \int x dx + \int x^{2} dx + ... = \sum_{i\geq 0} \int x^{i} dx. $$ Since the integral is distributed, you know you can just deal with the i-th term and have the correct result. – Bilbottom Jun 25 '16 at 08:49
  • @Bilbottom That is not correct. You are forgetting a limit! In fact, in general one cannot exchange integral and (infinite) sums because you have $$\int \lim_{n \to \infty} \sum_{i=0}^n x^i dx $$ Now, if you can exchange limit and integral, then you are correct, but that is not true in general. You need to justify the exchange of integral and series using either Fubini's theorem (hence looking at the series as an integral wrt the counting measure) or using Lebesgue dominated convergence theorem / monotone convergence theorem to exchange the limit and the integral. – Ant Jun 25 '16 at 08:55
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    @Ant I've not even come across those theorems - I guess my approach to infinite sums has been somewhat naive then. I'll give them a search and see if I can ammend my answer accordingly! – Bilbottom Jun 25 '16 at 09:17