Too big powers to integrate

Calculus Level 3

∫ 0 ∞ 1 1 + x 197 + x 2 + x 199 d x = a π b c \large\int_0^\infty\dfrac{1}{1+x^{197}+x^2+x^{199}}dx = \frac{a\pi^b}{c}

The equation above holds true for positive integers a a , b b and c c , where a a and c c are coprime integers. Find the value of ( a + b + c ) 2 + 3 (a+b+c)^2 + 3 .


The answer is 39.

A Former Brilliant Member by A Former Brilliant Member

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

Chew-Seong Cheong
May 18, 2018

Relevant wiki: Integration Tricks

I = ∫ 0 ∞ 1 1 + x 2 + x 197 + x 199 d x Using ∫ 0 ∞ f ( x ) d x = ∫ 0 ∞ f ( 1 x ) x 2 d x = 1 2 ∫ 0 ∞ ( 1 1 + x 2 + x 197 + x 199 + 1 x 2 ( 1 + x − 2 + x − 197 + x − 199 ) ) d x = 1 2 ∫ 0 ∞ ( 1 1 + x 2 + x 197 + x 199 + 1 x 2 + 1 + x − 195 + x − 197 ) d x = 1 2 ∫ 0 ∞ ( 1 1 + x 2 + x 197 + x 199 + x 197 x 199 + x 197 + x 2 + 1 ) d x = 1 2 ∫ 0 ∞ 1 + x 197 1 + x 2 + x 197 + x 199 d x = 1 2 ∫ 0 ∞ 1 1 + x 2 d x = tan ⁡ − 1 x 2 ∣ 0 ∞ = π 4 \begin{aligned} I & = \int_0^\infty \frac 1{1+x^2+x^{197}+x^{199}} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f \left(\frac 1x\right)}{x^2} \ dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{1+x^2+x^{197}+x^{199}} + \frac 1{x^2\left(1+x^{-2}+x^{-197}+x^{-199}\right)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{1+x^2+x^{197}+x^{199}} + \frac 1{x^2+1+x^{-195}+x^{-197}} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac 1{1+x^2+x^{197}+x^{199}} + \frac {x^{197}}{x^{199}+x^{197}+x^2+1} \right) dx \\ & = \frac 12 \int_0^\infty \frac {1+x^{197}}{1+x^2+x^{197}+x^{199}} dx \\ & = \frac 12 \int_0^\infty \frac 1{1+x^2} dx = \frac {\tan^{-1} x}2 \bigg|_0^\infty = \frac \pi 4 \end{aligned}

Therefore, ( a + b + c ) 2 + 3 = ( 1 + 1 + 4 ) 2 + 3 = 39 (a+b+c)^2+3 = (1+1+4)^2+3 = \boxed{39} .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

@Chew-Seong Cheong Sir I did using the method given below

I = ∫ 0 ∞ 1 1 + x 2 + x 197 + x 199 d x I = \large\int_0^\infty\dfrac{1}{1+x^2+x^{197}+x^{199}}dx

I = ∫ 0 ∞ 1 ( 1 + x 2 ) ( 1 + x 197 ) d x I = \large\int_0^\infty\dfrac{1}{(1+x^2)(1+x^{197})}dx

Let x x = = tan ⁡ θ \tan\theta

Therefore the integral reduces to

I = ∫ 0 π 2 sec ⁡ 2 θ ( sec ⁡ 2 θ ) ( 1 + tan ⁡ 197 θ ) d θ I = \large\int_0^\frac{\pi}{2}\dfrac{\sec^2\theta}{(\sec^2\theta)(1+\tan^{197}\theta)}d\theta

I = ∫ 0 π 2 cos ⁡ 197 θ sin ⁡ 197 θ + cos ⁡ 197 θ d θ I = \large\int_0^\frac{\pi}{2}\dfrac{\cos^{197}\theta}{\sin^{197}\theta+\cos^{197}\theta}d\theta

Using f ( x ) = f ( a + b − x ) f(x) = f(a+b-x)

I = 1 2 ∫ 0 π 2 d θ I= \dfrac{1}{2}\large\int_0^\frac{\pi}{2}d\theta

I = π 4 I = \large\dfrac{\pi}{4}

A Former Brilliant Member - 3 years ago

Log in to reply

Realy nice method!! Just instead of dx it should be d(theta) after the substitution step.

PRANAV Singla - 3 years ago

Log in to reply

Thanks, I have edited the solution

A Former Brilliant Member - 3 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...