# Relation between $\ln$ and $\arctan$

Let $$a>0$$ and consider the integral $$$$I(x) = \int_a^x \frac{1}{1+u} \, du.$$$$

On one hand, $$$$I(x) = \ln(1+x) - \ln(1+a) = \ln\left(\frac{1+x}{1+a}\right).$$$$

And on the other, if we consider $$u$$ as $$(\sqrt{u})^2 = |u| = u$$ on $$(a,x)$$, $$$$I(x) = \arctan(\sqrt{x}) - \arctan(\sqrt{a}).$$$$

This implies, $$$$\ln\left(\frac{1+x}{1+a}\right) = \arctan(\sqrt{x}) - \arctan(\sqrt{a}).$$$$

Which seems like an interesting enough equality to feel like I should have seen something about it before. I mean, it does not seem at all obvious that natural log and inverse tangent should be related. Or is it? Am I missing something? Is there an error in my logic?

• Have you tried testing that equality numerically with a few choices of $x$ and $a$? Dec 26, 2023 at 16:44
• Even without evaluating numerically, the left-hand-side goes to infinity as $x$ goes to infinity, while the right-hand-side is bounded. Dec 27, 2023 at 5:42

Your mistake is that in your arctan method you're asserting that $$\left(f(\sqrt{x})\right)' = f'(\sqrt{x})$$, since if you integrate both sides you get $$f(\sqrt{x}) = \int \left(f(\sqrt{x})\right)' = \int f'(\sqrt{x})$$ and taking $$f(x) = \arctan(x)$$ you get your second (erroneous) solution. This is incorrect, as the correct relation is $$\left(f(\sqrt{x})\right)' = f'(\sqrt{x})\cdot \frac{1}{2\sqrt{x}}$$ which is given by the chain rule. The way you undo the chain rule when integrating is doing a $$u$$-sub, which is (as others have pointed out) what you forgot to do.

Interestingly enough, there is a relationship between logarithms and the inverse tangent function, but it's a bit complex.

Defining the number $$i = \sqrt{-1}$$, then $$i$$ thus satisfies $$i^2= -1$$. Having this definition, by expanding the product as usual you can show that $$x^2 +1 = (x+i)(x-i)$$. Then, by partial fractions we get \begin{align*} \arctan(x)& = \int \frac{1}{x^2+1}\, \mathrm{d}x \\ &= \int \frac{1}{(x+i)(x-i)}\, \mathrm{d}x\\ &=\frac{i}{2} \int \frac{1}{x+i}\, \mathrm{d}x - \frac{i}{2} \int \frac{1}{x-i}\, \mathrm{d}x\\ & = \frac{i}{2}\ln(x+i) - \frac{i}{2}\ln(x-i) +C\\ & = \frac{i}{2}\ln\left(\frac{x+i}{x-i}\right)+C \end{align*} You can, in fact, find these kinds of logarithmic relations for all inverse trig functions, which at their core are possible because of Euler's formula $$e^{ix} = \cos(x) +i \sin(x)$$ which relates trig functions with exponentials, thus also relating their inverses.

• Remark: The expression $\arctan(x)$ is analgous to $\text{arctanh}(x)=\frac 12\log(\frac{1+x}{1-x})$. Dec 26, 2023 at 5:45
• A bit complex, is it? Dec 26, 2023 at 17:24

If you differentiate $$f(x) = \arctan(\sqrt x)$$ you will find that $$f'(x)$$ is not $$\frac{1}{1+x}$$ (because of the chain rule). So the second method is incorrect.

Your second method makes the substitution $$v=\sqrt u$$, so you should get $$du=2v dv\ne dv$$, hence the substitution would not yield an integrand of $$\dfrac{1}{1+v^2}$$.

$$I(x) = \int_a^x \frac{1}{1+u} \, du$$

Cross checking by differentiation of OP's second result: $$\frac{d}{du}(\arctan\sqrt u){=\frac{1}{1+(\sqrt u)^2}\cdot\frac{d}{du}(\sqrt u)\quad (\text{By chain rule})\\=\frac{2\sqrt u}{1+u}}\\ \therefore \int \frac{1}{1+u} \, du\ne \arctan \sqrt u +c \,\\ \text{And, } \int \frac{2\sqrt u}{1+u} \, du= \arctan \sqrt u +c \,$$

Explaining the correct way second process:

$$\sqrt{u}=:y\iff \frac{1}{2\sqrt{u}}du=dy\iff du=2ydy\\\int \frac{1}{1+u} du{=\int \frac{2ydy}{1+y^2}\\=\int \frac{d(1+y^2)}{1+y^2}\\=\ln|1+y^2|+c\\=\ln|1+u|+c}$$ Therefore, $$I(x) = \int_a^x \frac{1}{1+u}=\ln(1+x) - \ln(1+a) = \ln\left|\frac{1+x}{1+a}\right|$$

OP's mistake: Changing of variable.

The answer with the logarithm is correct.

For the answer you obtain with the arctangent, you want to write $$u = (\sqrt u)^2$$ so you may want to perform a change of variables $$t = \sqrt u$$. Redo your working and see if the answer you obtained is correct or not.