The Most Complicated And Beautiful Integral I Have Ever Solved…
Regardless of whether you are a mathematician or not, at first glance, this question just looks like someone has just forced in all kinds of series and complications. True mathematicians after looking at it for a minute or 2 will realize the simplicity in the question and the trick being played to fool all those who see it. This question can be solved by those who know how to solve integrals of course but along with that you should know the Taylor series for functions and how you expand them. A little bit of complex numbers and sequence and series is involved in it but those are basics which everyone should know if you know till the level of solving integrals.
For those who cannot make it out, I am here to solve this entire monstrosity in this post. Let’s go in step be step. Before I start, I would like to mention that one of my friends had shared this as a reel to me challenging me to solve it. I did do it and it just made my day. Credits to him for sharing this to me!
Let’s look at the lower limit of the given integral carefully, summation of (iπ)^n over n! up to infinity is just the short hand representation for e raised to iπ. and when we replace that summation with e raised to iπ we get the lower limit to be e raised to iπ + 1. Regardless of whether you know the Euler representation of complex numbers, if you truly love mathematics, you should know the value of this expression. I myself have talked a lot about this expression in my posts. The value of that is zero and that as a whole gives the lower limit to be zero.
Let’s go to the upper limit which actually looks a bit lengthy compared to the lower one. You should know that the sum of infinite terms of geometric progression is given by a=1/(1-r) where r is between 0 and 1 and a is the first term of the geometric progression. (A geometric progression is a sequence where you keep on multiplying the previous term with a constant number to get the next term of the series).
Taking this and solving the first term of the upper limit we get the answer pretty elegantly,
Now the next term of the upper limit is a very basic integral. Not much special in that and so here is the answer of that,
Both the terms gave us the final answer as π/4. So adding both we get the upper limit as π/2.
Before we get into the most scary and time taking part for many which is the integrand, let’s solve the power of the integrand to which it’s raised to first. I have talked about the factorial and the variations of it in one of my older posts, do check it out if you need to know about the factorial. Anyways, factorial is a kind of short hand representation of continuous product of all natural numbers until the number mentioned with the factorial (Including that as well). So 3! gives 1x2x3=6. And 4! is 24. Which means (3!-2)/(4!-³²) = 4/15.
The integral which is multiplies to this is also a very basic one and the solution for that is also here:
Look at the beauty! The power is 1!!!
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Our job is much easier now when we think about the final integral.
Now let’s get into the most dangerous and probably the only place where a vast majority might suffer. The integrand. You will be amazed to see how the final integral is going to look like after I solve the integrand. Already you would have understood from my solution of the limits and the power of the integrand that this question is not at all tough but only a trick being played to scare all those who try to encounter it. It really takes a lot of practice and love towards the subject to understand the category to which a question belongs to at first glance. (The right one lol)….
The 2 summations given to us are 2 types of trigonometric functions. For a lot of you who don’t have much knowledge about series expansions of functions, this might be surprising or inacceptable. But, there are methods through which a function can be expanded to a type of series to get it’s most appropriate value for corresponding inputs. Here,
We have got these from McLaurin expansion of these functions. For those who don’t have much idea about this, I am sorry to say that even I don’t have an in depth knowledge in this particular area. But I’ll surely talk about this in future posts when I get to learn about this. But in India, those students whoa re preparing for the JEE have a basic level of knowledge about this and that’s how even I recognized this.
Now the integrand is simply Tan(x) over (Tan(x)+Cot(x)). Now honestly anyone who knows basics of integration and trigonometry can solve this integral:
An alternative approach is the use of the famously known King’s property of integration: (Note: This is an unofficial name of the property given by Indian teachers and used by students in India. There is as such no officially recognized name for this property.)
Well, that’s a lot of mathematics for this post. My ultimate intention of solving this question here was to show those who think that bigger and complicated looking problems are tough are not always right. Crazily complex questions elegantly simplify to beautiful answer like this one. I hope you enjoyed this one. Once again, kudos to the guy who had forwarded this to me!
PS: I missed dx in the main integral in the image. Hope you don’t mind! 😁
Thank You!…
