How to read $\int_0^x \frac{1}{1+t^2}dt= \int_0^1 ( 1-t^2+ \cdot \cdot \cdot + (-1)^t t^{2n})dt$?

Question

How to read $$\int_0^x \frac{1}{1+t^2}dt= \int_0^1 ( 1-t^2+ \cdot \cdot \cdot + (-1)^t t^{2n})dt$$

Specifically, why the integral switches from being $\int_0^x$ to being $\int_0^1$? Is this because $\int_0^x$ is "integral over all $x$" and since the series expansion for $\frac{1}{1+x^2}$ is defined for $|x|<1$, then one can get "all $x$" by integrating only the interval [0,1]?

However, since $\frac{1}{1+x^2}$ is surely defined for $[-1,1]$, then shouldn't the above rather be

$$2 \cdot \int_0^1 ( 1-t^2+ \cdot \cdot \cdot + (-1)^t t^{2n})dt$$

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This appears in p.11 of this document.

Answer 1

Clearly the upper bound of integration should be the same. This is a typo.

Moreover the two expressions are not equal: in the second there should be the error term appearing at the very end for these to be equal, which is then bounded just below.

In formulas:

$$\int_0^x \frac{1}{1+t^2}dt= \int_0^x( 1-t^2+ \cdot \cdot \cdot + (-1)^t t^{2n})dt+R_n(x)$$

Answer 2

$$\frac{1}{1 + t^2}$$

Is the result of the geometric series

$$\sum_{k = 0}^{+\infty} (-x)^k$$

Only when $|x| < 1$.

Your integration is bounded from $0$ to $x$, where $x$ is unknown, so it might be $1$, $10$ or $10^{44}$. Since you don't know what $x$ is, you have either to split the integration into two parts like

$$\int_0^1 + \int_1^x$$

Or use the geometric series only by supposing $x = 1$. If you suppose that, then you are legitimate to use the geometric expansion to get

$$\int_0^1 \frac{1}{1 + t^2}\ \text{d}t = \int_0^1 \sum_{k = 0}^{+\infty} (-t)^k\ \text{d}t = \sum_{k = 0}^{+\infty} (-1)^k\ \int_0^1 t^k\ \text{d}t$$

The integral becomes trivial:

$$\sum_{k = 0}^{+\infty} \frac{(-1)^k\ x^{k+1}}{k+1}$$

Which is exactly the expansion of the arcotangent.

On the other side, the function $\frac{1}{1+t^2}$ is an even function, so you can indeed write this:

$$\int_0^1 \frac{1}{1 + t^2}\ \text{d}t = \frac{1}{2}\int_{-1}^{+1} \frac{1}{1 + t^2}\ \text{d}t$$

and act by consequence.

Your error is the $2$. You need $\frac{1}{2}$ since you're doubling the integral over the range $[-1, +1]$, hence by doubling it, you need the halfpart.