Time for our next problem in the POW series! Earlier, Todd and I deliberated for a bit on whether we should pose a “hard” Ramanujan identity (involving an integral and Gamma function) as the next POW, but decided against doing it. Perhaps, we may do so some time in the future.
Okay, the following integral was brought to our attention by Carl Lira, and for the time being I won’t reveal the actual source of the problem.
Compute 
It is “hard” or “easy” depending on how you look at it!
Please send your solutions to topological[dot]musings[At]gmail[dot]com by Wednesday, June 26, 11:59pm (UTC); do not submit solutions in Comments. Everyone with a correct solution gets entered in our Hall of Fame! We look forward to your response.
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June 20, 2008 at 5:54 pm
Paul
You haven’t indicated limits, so this is an indefinite integral right? We are to compute the integral as a function of x?
June 20, 2008 at 5:59 pm
Vishal Lama
Yes, this is an an indefinite integral.
June 25, 2008 at 8:37 am
Dave S
I think we can find its analytical integration.
June 27, 2008 at 12:59 am
Solution to POW-5: A “hard” integral! « Todd and Vishal’s blog
[…] Week (POW), Puzzles | Tags: hard integral | by Vishal Lama We got some very good response to our last week’s problem from several of our “regular” problem-solvers as well as several others who are […]
August 17, 2008 at 3:40 pm
Anonymous
Here’s my solution to the
The hard integral above:
∫(x²-1)/((x²+1)*√(x^4+1))dx
First divide both (x² – 1) and (x² + 1) by x, then factor out an x from √(x^4 + 1). This results in the equivalent form
∫ (x – (1/x)) dx / [x (x + (1/x)) √(x² + (1/x²))]
Next,
x² + (1/x²) = x² + 2 + (1/x²) – 2
= (x + (1/x))² – 2
Inserting this, we get
∫ (x – (1/x)) dx / [x (x + (1/x)) √{(x + (1/x))² – 2}]
let x = e^u; dx = e^u du, and the integral becomes
∫ (e^u – e^(-u)) du / [(e^u + e^(-u)) √{(e^u + e^(-u))² – 2}]
Next, another substitution:
v = e^u + e^(-u)
dv = (e^u – e^(-u)) du
This gives
∫ dv/[v √(v² – 2)]
Yet another substitution:
v = √2 sec(θ)
√(v² – 2) = √[2 (sec²(θ) – 1)] = (√2) tan(θ)
dv = √2 sec(θ) tan(θ) dθ
which yields
∫ √2 sec(θ) tan(θ) dθ / [√2 sec(θ)(√2 tan(θ))]
= ∫ (1/√2) dθ
= (θ/√2) + C
Now, back-substituting,
v = √2 sec(θ)⇒ θ = arcsec(v/√2) / √2
But v = e^u + e^(-u) = x + (1/x)
So in terms of original variable x, the integral is
arcsec([x + (1/x)]/√2) / √2 + C
October 4, 2008 at 2:00 pm
POW-10: Another hard integral? « Todd and Vishal’s blog
[…] problem of the week has been long overdue. Here’s one that may appeal to those who liked POW-5 (maybe we should have an integration problem every fifth time?): […]