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Motivated by the beautiful identity

$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$$

and the equally beautiful proof of

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$

that it implies (the proof is presented in this post), I decided to try and find a closed-form expression for $\int_{0}^{\pi/4}\tan^{2n+1}(x)dx$. Following the method I used to establish the integral above (reduction formulae), I discovered this stunning identity, true for all $n\geq 1$ (also true for $n=0$ if you accept the empty sum convention).

$$\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$$

As you can probably see, this is very suggestive of the fact that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$, so I naturally tried to leverage it in a proof. Here's my attempt:

Proof: We'll begin by proving that $\int_{0}^{\pi/4}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$. We do this with mathematical induction.

The base case $n=1$ can be established as follows:

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2(1)+1}(x)dx &= \int_{0}^{\frac{\pi}{4}}\tan^{3}(x)\text{ }dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan(x)\left[\sec^2(x)-1\right]dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan(x)\text{ }dx \end{align*}

Making the substitution $u=\tan(x)$ in the first integral gives

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan(x)\text{ }dx &= \int_{0}^{1}u\text{ }du-\left[\ln\left(\sec\frac{\pi}{4}\right)-\ln(\sec 0)\right]\\ &= \frac{1}{2}-\ln(\sqrt{2})\\ &= -\left(\frac{1}{2}\ln(2)-\frac{1}{2}\cdot 1\right)\\ &= (-1)^1\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{1}\frac{(-1)^{k-1}}{k}\right) \end{align*}

This proves the base case. If we assume that the identity is true for some $m\in\mathbb{N}$, then

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2(m+1)+1}(x)dx &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1+2}(x)\text{ }dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\tan^{2}(x)\text{ }dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\left[\sec^2(x)-1\right]dx\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\text{ }dx \end{align*}

Like before we make the substitution $u=\tan(x)$ in the first integral. Leveraging our assumption, we get

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\text{ }dx &= \int_{0}^{1}u^{2m+1}\text{ }du-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\ &= \frac{1}{2m+2}-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\ &= \frac{(-1)^m(-1)^m}{2(m+1)}-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\ &= (-1)^m\left[\frac{(-1)^m}{2(m+1)}-\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\right]\\ &= (-1)^m\cdot -\left(-\frac{(-1)^{m}}{2(m+1)}+\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\ &= (-1)^{m+1}\left(\frac{1}{2}\ln(2)-\frac{(-1)^{(m+1)-1}}{2(m+1)}-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\ &= (-1)^{m+1}\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m+1}\frac{(-1)^{k-1}}{k}\right) \end{align*}

This shows that it must be true for $m+1$. Thus, mathematical induction implies that the identity is true for all natural numbers.

We are now in a position to prove that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$. Notice that

\begin{align*} \left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right| &= \left|(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)\right|\\ &= \left|\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx\right|\\ &= \int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx \end{align*}

since $\tan^{2n+1}(x)\geq 0$ for $0\leq x <\pi/2$. We can bound the quantity

$$\left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|$$

by making the substitution $x=\tan^{-1}(t)$ on the last integral and proceeding as follows

\begin{align*} \int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx &= \int_{0}^{1}\frac{t^{2n+1}}{1+t^2}dt\\ &< \int_{0}^{1}t^{2n+1}\text{ }dt\\ &= \frac{1}{2n+2}\\ &= \frac{1}{2(n+1)} \end{align*}

This gives the upper bound $\left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|<\frac{1}{2(n+1)}$, or

$$\left|\ln(2)-\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|<\frac{1}{n+1}$$

Letting $n\to\infty$ and applying the squeeze theorem immediately gives the desired result.

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$$

Q.E.D.

If correct, this argument is absolutely beautiful! Not only does it prove the equality

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$$

in a completely elementary fashion, it also establishes this absolutely astounding result:

$$\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$$

In particular, combining this with $\int_{0}^{\pi/4}\tan^{2n}(x)\text{ }dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$, we now have a recipe for evaluating

$$\int_{0}^{\frac{\pi}{4}}\tan^{n}(x)\text{ }dx$$

for any natural number $n$. How awesome it that?!

Let me know what you think! If you have any suggestions for optimizing the proof or identify some logical errors, please don't hesitate to share them with me!

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    $\begingroup$ Using a Taylor series for $\ln{(1+x)}$ seems more direct. $\endgroup$ – user58697 Mar 20 at 22:04
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    $\begingroup$ @user58697 The Taylor series expansion only holds for $|x|<1$, though. Proving that it holds for the endpoints requires separate arguments, hence the proof I gave. $\endgroup$ – Alann Rosas Mar 20 at 22:05
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    $\begingroup$ The two formulas can be found as item 3.622 in Gradshteyn and Ryzhik (page 370 in the 1980 edition I have) $\endgroup$ – Jean Marie Mar 20 at 23:36
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    $\begingroup$ Let me say that I have upvoted your question because the ideas you have and the enthousiasm you show will be helpful in your studies... $\endgroup$ – Jean Marie Mar 20 at 23:39
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    $\begingroup$ @JeanMarie thank you for the kind words and book mention! $\endgroup$ – Alann Rosas Mar 21 at 2:53

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