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 ...

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.

Last edited:

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.

"How to integrate with functions of differentials?"

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