Problem
For x ∈ R, let tan⁻¹x ∈ (-π/2, π/2). Then, find the minimum value of the function f: R → R, f(x) = \( \int_{0}^{x\tan^{-1}(x)}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \)
Solution
A disturbing integral. Thankfully, we don’t need to evaluate it. To find its minimum value, we’ll look at two different methods.
Method 1
Let’s start by differentiating f(x) w.r.t. x, and analyse its derivative. We’ll use the Leibniz rule.
We’ll get f'(x) = \( \frac{e^{x\tan^{-1}x-\cos{(xtan^{-1}x)}}}{1+({xtan^{-1}x})^{2023}}\left(\frac{x}{1+x^2}+tan^{-1}x\right) \)
Eww. But let’s not lose hope yet.
If you take a closer look at the scary function outside the bracket, the numerator is a power of e, which is a positive number. And the denominator is positive as well, since xtan⁻¹x is always positive.
That means, the entire function is positive, and we dont need to worry about it.
Let’s make f'(x) look nicer: f'(x) = g(x)\(\left(\frac{x}{1+x^2}+tan^{-1}x\right) \), where g(x) is the scary function.
Now, the function inside the bracket is negative for x < 0, positive for x > 0, and zero for x = 0.
That means, f(x) is decreasing for x < 0, increasing for x > 0, and attains its minimum value at x = 0.
Finally, the minimum value of f(x) equals f(0) or \( \int_{0}^{0}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \) which equals 0.
Method 2
Let’s take a look at f(x) again: \( \int_{0}^{x\tan^{-1}(x)}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \)
Consider the function inside the integral, i.e. g(t) = \( \frac{e^{t-\cos{t}}}{1+t^{2023}}dt \) – it is always positive for t > 0 (for reasons described above).
That means, if we integrate g(t) from 0 to a positive number x, the resulting function will be positive and will increase with increasing values of x.
Why? Think in terms of area – if the function is positive, then the area bounded by the function with the x-axis is also positive. And, with increasing values of x, the bounded area keeps on increasing.
So, \( \int_{0}^{x}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \) is an increasing function. But we had xtan⁻¹x instead of x as the upper limit.
Since xtan⁻¹x is also always positive for x > 0, and increases with increasing values of x, \( \int_{0}^{x\tan^{-1}(x)}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \) will be also be an increasing function.
What happens when x < 0? Note that f(-x) = \( \int_{0}^{-x\tan^{-1}(-x)}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt = \int_{0}^{x\tan^{-1}(x)}\frac{e^{t-\cos{t}}}{1+t^{2023}}dt \) = f(x).
That is, f(x) is an even function, or symmetric about the y-axis.
That means, f(x) is decreasing when x < 0 and increasing when x > 0. So, when will it attain its minimum value? At x = 0.
So, the minimum value is f(0) which equals 0, as we calculated in the first method.
Comments
An ugly looking integral was given to scare away students. But the integral wasn’t supposed to be calculated. Using either of the methods quickly takes you to the conclusion that the function attains its minimum value at x = 0. The second method is a bit more clever, and saves the differentiation effort as well. Overall, a fairly easy problem, given that you don’t get scared.
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