The Wicked Integral

Calculus Level 5

∫ 0 ∞ d x x [ x 2 + ( 1 + 2 2 ) x + 1 ] [ 1 − x + x 2 − x 3 + ⋯ + x 2014 ] {\Large\int_0^\infty} \frac{dx}{\sqrt{x} \left[x^2+\left(1+2\sqrt{2}\right)x+1\bigg] \bigg[1-x+x^2-x^3+\cdots+x^{2014}\right]}

Given that the integral above is equal to π ( a − b ) \pi(\sqrt a - b) , where a a and b b are positive integers, find the value of a + b a+b .


The answer is 3.

View wiki
Anastasiya Romanova by Anastasiya Romanova , 20, Russia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Shivang Jindal
Oct 23, 2014

Substituting , x = 1 / u x= 1/u and adding the two integrals, gives,

I = ∫ 0 ∞ x 2 + 1 x 4 + ( 1 + 2 2 ) x 2 + 1 d x I = \int_{0}^{\infty} \frac{x^2+1}{x^4+(1+2\sqrt{2})x^2+1} dx Last integral can be easily calculated by dividing numerator and denominator by x 2 x^2 .

This on simplification gives, ( 2 − 1 ) π (\sqrt{2}-1)\pi !

Very nice problem!

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Really nice problem! ⌣ ¨ \ddot\smile

Karthik Kannan - 6 years, 6 months ago

Hello Sir, Can you describe your steps of calculations please, I don't really understand how you came to that result properly! Thank you have a good day!

Seidemann Frost - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...