Animated Solution for Mathematics - Definite Integration: Let $f:(0,2)\to \mathbb{R}$ be defined as $f(x)={\log }_2\left(1+\tan \left(\frac{\pi x}{4}\right)\right)$. Then, ${\lim }_{n\to \infty }\frac{2}{n}\left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots +f(1)\right)$ is equal to ______

Imagine you are staring at this problem on your JEE Advanced paper. You see a limit as $n$ approaches infinity, a summation, a logarithm, and a tangent function all tangled together. It looks like a nightmare of complexity.

But here is the secret: in the world of competitive mathematics, complexity is often just a mask for elegance. Let us peel back that mask together.

We are given the expression:

At first glance, you might be tempted to try and expand the sum or look for a pattern in the terms. Resist that urge! The presence of $\frac{1}{n}$ outside the sum and $\frac{r}{n}$ inside the function is a massive, flashing neon sign pointing toward the Riemann Sum definition.

Recall the fundamental definition of a definite integral as the limit of a Riemann sum:

In our expression, we have a factor of $\frac{2}{n}$. We can simply pull the constant $2$ outside the limit.

Now, the remaining part $\frac{1}{n}\sum f(\frac{r}{n})$ maps perfectly to the integral definition. Our discrete variable $\frac{r}{n}$ transforms into the continuous variable $x$, and the $\frac{1}{n}$ becomes our differential $dx$.

The limits of integration, as $r$ goes from $1$ to $n$, naturally span from $0$ to $1$. Suddenly, our infinite sum has collapsed into a clean, definite integral:

Now, look at the integrand. Can you integrate ${\log }_2(1+\tan (\frac{\pi x}{4}))$ directly? It is not impossible, but it is certainly not straightforward.

This is where we invoke the 'King' of definite integration properties:

This property is your best friend when dealing with trigonometric functions in definite integrals. Let us apply it here, where $a=1$. We replace every $x$ with $(1-x)$:

Now, we must simplify the argument of the tangent function. Distributing the $\frac{\pi }{4}$, we get $\frac{\pi }{4}-\frac{\pi x}{4}$.

We use the trigonometric identity for the tangent of a difference, $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$. Since $\tan (\frac{\pi }{4})=1$, the expression simplifies beautifully:

Now, substitute this back into our logarithm argument:

Look at what we have achieved! Our integral now looks like this:

Using the quotient rule for logarithms, $\log (\frac{a}{b})=\log a-\log b$, we can split this into two parts:

Since ${\log }_{2}2=1$, we can distribute the integral:

The second term is exactly our original integral $E$! We have arrived at the equation:

Solving for $E$ is trivial now: $2E=2$, which means $E=1$. What started as a terrifying infinite sum has been reduced to the integer $1$ through the sheer power of symmetry and algebraic manipulation.