# I How to integrate with functions of differentials?

#### jk22

I fell upon such a wrting :

$$du=tan(d\theta)$$

How to integrate this ?

I didn't try numerically but I thought of expanding the tangeant in series but then should for example $$\int d\theta^2$$ be understood as a double integration ?

#### BvU

Homework Helper
No, $d\theta^2 = 2\theta d\theta$.

You fell ? Did it hurt ? Was there a context that can help us make it clear to you or did it get lost in the fall ?

#### Delta2

Homework Helper
Gold Member
No, $d\theta^2 = 2\theta d\theta$.
it depends whether we mean $d(\theta^2)$ or $(d\theta)^2$

And for me to answer the OP's question $\int (d\theta)^n$ is zero for all $n\geq 2$

• BvU

#### BvU

Homework Helper
@Delta2 corrected me rightly: the Taylor series at $\theta = 0$ contains $(d\theta)^n$, not $d\theta^2$.

Still curious about the context ...

• Delta2

#### jk22

Yes missed parentheses from my part.

The context is easy :

Let be the parametrization of a unit sphere $\vec{r}(\theta,\phi)=(cos\theta cos\phi,cos\theta sin\phi,sin\theta),\theta\in [-\pi/2,\pi/2],\phi\in [0,2\pi]$ and the associated basis vectors $\vec{e}_\theta=\frac{\partial\vec{r}}{\partial\theta}$ and $\vec{e}_\phi$

The metric is $g=diag(1,cos(\theta)^2)$

I wanted to orthonormalize the metric via the associated quadratic form : $\underbrace{(du\vec{e}_\theta+dv\vec{e}_\phi)}_{\vec{x}}g\vec{x}^T$

The $du,dv$ are infinitesimals on the tangent plane and I express them as a function of $\theta,\phi$: $du=tan(d\theta)$
$dv'=cos(\theta)^2tan(d\phi)$

Instead of linearizing I did what I wrote in the OP with successive integrations ($\int(d\theta)^n=\theta^n/n!$ by fixing all the successive constants of integration to 0) and found
$u=\sum_{n=0}^\infty(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}\frac{\theta^{2n+1}}{(2n+2)!(2n+1)!}$

I wanted to compute the shape of u,v which if you want should be a map of the sphere.

Is it periodic and what's the asymptotic behaviour ?

Could anyone help me or give me references in webpages, it should be found in common textbooks about differential geometry or map projections I suppose.

What I found weird is that if I consider a path on the sphere $\theta(t),\phi(t)$ then the endpoint coordinate on the map could depend on the path itself ?! Thanks.

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#### archaic

I think $du=\frac{d(\tan{x})}{dx}$ at 0: $d(\tan{x})=(\tan{(0+dx)}-\tan{0})dx$. So perhaps integrating this would be $\int du \cdot dx=\int (1+\tan^2{0})dx$?

#### jk22

$du$ is an infinitesimal while $\frac{d(\tan(x))}{dx}$ is a finite quantity if $|x|<\frac{\pi}{2}$. Did you mean $du=d\tan(\theta=x)=(1+\tan(\theta)^2)d\theta$, so that we have not $\tan(d\theta)$, which is meaningless ?

I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.

#### Mark44

Mentor
$du$ is an infinitesimal while $\frac{d(\tan(x))}{dx}$ is a finite quantity if $|x|<\frac{\pi}{2}$.
Yes, that is more-or-less correct. The expression $\frac{d(\tan(x))}{dx}$ is the derivative of the tangent function, with respect to its variable x.

The substitution was $u = \tan(x)$ from which the differentials of the two sides would be as you have below. Note that we would usually write $\tan^2(\theta)$ rather than $\tan(\theta)^2$.

However, I'm not sure this is relevant to your problem where you had $du = \tan(d\theta)$. I would be interested in seeing the work that led up to this equation.
jk22 said:
Did you mean $du=d\tan(\theta=x)=(1+\tan(\theta)^2)d\theta$, so that we have not $\tan(d\theta)$, which is meaningless ?

I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.

#### jk22

However, I'm not sure this is relevant to your problem where you had $du = \tan(d\theta)$. I would be interested in seeing the work that led up to this equation.
In fact there is no work : I suppose a circle and the tangent line at coordinate $\theta$, then an infinitesimal change $d\theta$ leads to a change $du=\tan(d\theta)$ on this line, with just basic trigonometrical geometry.

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