# How to prove the definition of arctangent by G. H. Hardy through integral?(update)

1. Oct 16, 2012

### zhongbeyond

1. The problem statement, all variables and given/known data
From introduction to analysis,by Arthur P. Mattuck,problem 20-1.

Problems 20-1
One way of rigorously defining the trigonometric functions is to start with the definition of the arctangent function. (This is the route used for example in the classic text Pure Mathematics by G. H. Hardy.)
So, assume amnesia has wiped out the trigonometric functions (but the rest of your knowledge of analysis is intact). Define
$$T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}$$

(a) Prove T(x) is defined for all x and odd.
(b) Prove T(x) is continuous and differentiable, and find T(x).
(c) Prove T(x) is strictly increasing for all x; find where it is convex, where
concave, and its points of inflection.
(d) Show T(x) is bounded for all x, and |T(x)| < 2.5, using comparison
of integrals. Can you get a better bound?

2. Relevant equations

3. The attempt at a solution

2. Oct 16, 2012

### Ray Vickson

Re: How to prove the definition of arctangent by G. H. Hardy through integral?(update

I don't know where the 2.5 arises; I get instead 1.875, by using 1/(1+t^2) < 1/t^2 for t >= 1 and by choosing an appropriate, simple g(t) such that 1/(1+t^2) <= g(t) on [0,1].

RGV

3. Oct 17, 2012

### zhongbeyond

Re: How to prove the definition of arctangent by G. H. Hardy through integral?(update

Thanks for replying,I finally solve it through help of dr.math(http://mathforum.org/dr.math/)

Break the inteval [0,x] to [0,a] + [a,x].
$$T(x)=\int_{0}^{x}\frac{dt}{1+t^{2}}= \int_{0}^{a}\frac{dt}{1+t^{2}}+\int_{a}^{x}\frac{dt}{1+t^{2}}$$

For the first inteval [0,a], $$\frac{1}{1+t^{2}} <= 1$$
For the second inteval [a,x], $$\frac{1}{1+t^{2}} <= \frac{1}{t^{2}}$$

and change the value of a will get the magic value 2.5

4. Oct 17, 2012

### Ray Vickson

Re: How to prove the definition of arctangent by G. H. Hardy through integral?(update

OK, but the way I got 1.875 was to use 1/(1+t^2) <= g(t) on [0,1], where g(t) = 1 for 0 <= t <= 1/2 and g(t) = 3/2 - t for 1/2 < t <= 1 (and 1/t^2 for t > 1) The integral of g(t) over [0,1] is 7/8.

RGV