Complex Integral

Calculus Level 5

∫ 0 ∞ cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x \large \displaystyle \int _{ 0 }^{ \infty }{ \dfrac { \cos\left( 6{ x }^{ 2 } \right) -\sin\left( 6{ x }^{ 2 } \right) }{ 1+{ x }^{ 4 } } \, dx }

The integral above is equal to A π e − B C D \dfrac { A\pi { e }^{ -B } }{ C\sqrt { D } } , where A A and C C are positive coprime integers, B B is an integer and D D is a square-free integer.

Find A + B + C + D A+B+C+D .

For more calculus problems see this.


The answer is 11.

Aditya Kumar by Aditya Kumar , 22, India

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2 solutions

Mark Hennings
Jan 25, 2016

For any R > 1 R > 1 , integrate the holomorphic function f ( z ) = e 6 i z 2 1 + z 4 f(z) \,=\, \displaystyle\frac{e^{6iz^2}}{1+z^4} around the "quadrant contour" Γ R = γ 1 , R + γ 2 , R − γ 3 , R \Gamma_R \,=\, \gamma_{1,R} + \gamma_{2,R} - \gamma_{3,R} , where

  • γ 1 , R \gamma_{1,R} is the straight line segment from 0 0 to R R along the positive real axis,
  • γ 2 , R \gamma_{2,R} is the quarter circle arc given by z = R e i θ z = Re^{i\theta} for 0 ≤ θ ≤ 1 2 π 0 \le \theta \le \tfrac12\pi ,
  • γ 3 , R \gamma_{3,R} is the straight line segment from 0 0 to i R iR along the positive imaginary axis.

Then ∫ γ 1 , R f ( z ) d z = ∫ 0 R e 6 i x 2 1 + x 4 d x ∫ γ 3 , R f ( z ) d z = i ∫ 0 R e − 6 i x 2 1 + x 4 d x ( ∫ γ 1 , R − ∫ γ 3 , R ) f ( z ) d z = ∫ 0 R e 6 i x 2 − i e − 6 i x 2 1 + x 4 d x = ( 1 − i ) ∫ 0 R cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x \begin{array}{rcl} \displaystyle \int_{\gamma_{1,R}} f(z)\,dz & = & \displaystyle \int_0^R \frac{e^{6ix^2}}{1+x^4}\,dx \\ \displaystyle \int_{\gamma_{3,R}} f(z)\,dz & = & \displaystyle i\int_0^R \frac{e^{-6ix^2}}{1 + x^4}\,dx \\ \displaystyle \left(\int_{\gamma_{1,R}} - \int_{\gamma_{3,R}}\right) f(z)\,dz & = & \displaystyle \int_0^R \frac{e^{6ix^2} - ie^{-6ix^2}}{1+x^4}\,dx \; =\; (1 - i)\int_0^R \frac{\cos(6x^2) - \sin(6x^2)}{1+x^4}\,dx \end{array} Now z 2 z^2 has nonnegative imaginary part for any z ∈ γ 2 , R z \in \gamma_{2,R} , so that ∣ e 6 i z 2 ∣ ≤ 1 \big|e^{6iz^2}\big| \le 1 for all z ∈ γ 2 , R z \in \gamma_{2,R} ; this implies that ∫ γ 2 , R f ( z ) d z = O ( R − 3 ) R → ∞ . \int_{\gamma_{2,R}} f(z)\,dz \; = \; O(R^{-3}) \qquad \qquad R \to \infty \;. The only singularity of f ( z ) f(z) inside the contour Γ R \Gamma_R is a simple pole at 1 2 ( 1 + i ) \tfrac{1}{\sqrt{2}}(1+i) , and so ∫ Γ R f ( z ) d z = 2 π i R e s z = 1 2 ( 1 + i ) f ( z ) \int_{\Gamma_R} f(z)\,dz \; = \; 2\pi i \mathrm{Res}_{z=\tfrac{1}{\sqrt{2}}(1+i)} f(z) for any R > 1 R > 1 ; we deduce on letting R → ∞ R \to \infty that ( 1 − i ) ∫ 0 ∞ cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x = 2 π i R e s z = 1 2 ( 1 + i ) f ( z ) = 2 π i e − 6 2 i × 2 ( 1 + i ) = π e − 6 2 ( 1 + i ) ∫ 0 ∞ cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x = π e − 6 2 2 \begin{array}{rcl} \displaystyle (1-i)\int_0^\infty \frac{\cos(6x^2) - \sin(6x^2)}{1+x^4}\,dx & = & \displaystyle 2\pi i \mathrm{Res}_{z = \frac{1}{\sqrt{2}}(1+i)} f(z) \\ & =& \displaystyle 2\pi i \frac{e^{-6}}{2i \times \sqrt{2}(1+i)} \; = \; \frac{\pi e^{-6}}{\sqrt{2}(1+i)} \\ \displaystyle \int_0^\infty \frac{\cos(6x^2) - \sin(6x^2)}{1+x^4}\,dx & = & \displaystyle\frac{\pi e^{-6}}{2\sqrt{2}} \end{array} so that the answer is 1 + 6 + 2 + 2 = 11 1 + 6 + 2 + 2 \,=\, \boxed{11} .

9 Helpful 0 Interesting 0 Brilliant 0 Confused

My method was much shorter. It involved Feynman's method of integration and then solving a differential equation.

Aditya Kumar - 5 years, 4 months ago

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This method is a one-idea pony. If I had just written: integrating e 6 i z 2 1 + z 4 \frac{e^{6iz^2}}{1+z^4} around the upper-right quadrant (the integral along the quarter circle at infinity vanishes) gives ( 1 − i ) ∫ 0 ∞ cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x = ∫ 0 ∞ e 6 i x 2 − i e − 6 i x 2 1 + x 4 d x = 2 π i R e s z = 1 2 ( 1 + i ) f ( z ) ∫ 0 ∞ cos ⁡ ( 6 x 2 ) − sin ⁡ ( 6 x 2 ) 1 + x 4 d x = π e − 6 2 2 \begin{array}{rcl} \displaystyle (1 - i)\int_0^\infty \frac{\cos(6x^2) - \sin(6x^2)}{1+x^4}\,dx \; = \; \int_0^\infty \frac{e^{6ix^2} - ie^{-6ix^2}}{1+x^4}\,dx & = & \displaystyle 2\pi i \mathrm{Res}_{z = \frac{1}{\sqrt{2}}(1+i)} f(z) \\ \displaystyle \int_0^\infty \frac{\cos(6x^2) - \sin(6x^2)}{1+x^4}\,dx & = & \displaystyle\frac{\pi e^{-6}}{2\sqrt{2}} \end{array} I would not have done any less maths - just omitted some details.

Mark Hennings - 5 years, 4 months ago

@Aditya Kumar can u post an overview of your solution.

Seong Ro - 5 years, 4 months ago

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I will post the whole solution after 1hr.

Aditya Kumar - 5 years, 4 months ago
Aditya Kumar
Jan 26, 2016

Consider:

f ( p ) = ∫ 0 ∞ c o s ( p x 2 ) − s i n ( p x 2 ) 1 + x 4 d x \displaystyle f\left( p \right) =\int _{ 0 }^{ \infty }{ \frac { cos\left( p{ x }^{ 2 } \right) -sin\left( p{ x }^{ 2 } \right) }{ 1+{ x }^{ 4 } } dx }

Now on double differentiating it wrt p,

f ′ ′ ( p ) = − ∫ 0 ∞ x 4 c o s ( p x 2 ) − s i n ( p x 2 ) 1 + x 4 d x \displaystyle f''\left( p \right) =-\int _{ 0 }^{ \infty }{{ x }^{ 4 }\frac { cos\left( p{ x }^{ 2 } \right) -sin\left( p{ x }^{ 2 } \right) }{ 1+{ x }^{ 4 } } dx }

Now,

f ( p ) − f ′ ′ ( p ) = ∫ 0 ∞ c o s ( p x 2 ) − s i n ( p x 2 ) d x \displaystyle f\left( p \right) -f''\left( p \right) =\int _{ 0 }^{ \infty }{ cos\left( p{ x }^{ 2 } \right) -sin\left( p{ x }^{ 2 } \right) dx }

This clearly is a fresnel integral.

∴ f ( p ) − f ′ ′ ( p ) = 0 \displaystyle \therefore \quad f\left( p \right) -f''\left( p \right) =0

On solving the differential equation, we get:

f ( p ) = e − p C f\left( p \right) ={ e }^{ -p }C , where C is any constant.

Now for evaluating that constant,

f ( 0 ) = π 2 2 f\left( 0 \right) =\frac { \pi }{ 2\sqrt { 2 } }

Therefore, f ( p ) = π e − p 2 2 f\left( p \right) =\frac { \pi { e }^{ -p } }{ 2\sqrt { 2 } }

Substitute p=6 and we get the desired answer.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

The problem with this is that the theory which permits differentiation within the integral sign gets rather delicate if the integrands are not properly Lebesgue integrable (and not just improperly Riemann integrable). Your formula f ′ ′ ( p ) = − ∫ 0 ∞ x 4 cos ⁡ ( p x 2 ) − sin ⁡ ( p x 2 ) 1 + x 4 d x f''(p) \; = \; -\int_0^\infty x^4 \frac{\cos (px^2) - \sin(px^2)}{1+x^4}\,dx is contentious (note that x 4 x^4 needs to be inside the integral), since the integral only exists improperly. Certainly the improper integral exists, but is it the second derivative of f f ? Improper Riemann integrals are extremely delicate when it comes to reversing the order of limits in areas like differentiating under the integral sign.

If you defined f R ( p ) = ∫ 0 R cos ⁡ ( p x 2 ) − sin ⁡ ( p x 2 ) 1 + x 4 d x f_R(p) \; =\; \int_0^R \frac{\cos (px^2) - \sin(px^2)}{1 + x^4}\,dx then it would be true to say that f R ( p ) − f R ′ ′ ( p ) = F R ( p ) = ∫ 0 R ( cos ⁡ ( p x 2 ) − sin ⁡ ( p x 2 ) ) d x , f_R(p) - f_R''(p) \; = \; F_R(p) \; = \; \int_0^R \big(\cos (px^2) -\sin(px^2)\big)\,dx \;, and the RHS can be expressed in terms of the Fresnel integrals. You would need to solve this differential equation, and consider the limit of the solution as R → ∞ R \to \infty . At first sight, letting R → ∞ R \to \infty seems to let us replace f R f_R by 0 0 as you wish, but it is not easy!

The sort of formal calculation with improper integrals you are doing is suggestive, but would need a lot of tidying up to make a proof.

Mark Hennings - 5 years, 4 months ago

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Sorry for the typo.

Even I'm working on a proof for the Fresnel integral.

Aditya Kumar - 5 years, 4 months ago

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With F R ( p ) F_R(p) as in my previous post, the general solution of the differential equation f R ( p ) − f R ′ ′ ( p ) = F R ( p ) , f_R(p) - f_R''(p) \; = \; F_R(p) \;, is f R ( p ) = A e p + B e − p − ∫ 0 p F R ( u ) sinh ⁡ ( p − u ) d u . f_R(p) \; = \; Ae^p + Be^{-p} - \int_0^p F_R(u) \sinh(p-u)\,du \;. for some constants A , B A,B . Note that f R ′ ( p ) = A e p − B e − p − ∫ 0 p F R ( u ) cosh ⁡ ( p − u ) d u , f_R'(p) \; = \; Ae^p - Be^{-p} - \int_0^p F_R(u) \cosh(p-u)\,du \;, The function F R F_R is sufficiently well-behaved (see below) for this differentiation to be OK. Thus f R ( 0 ) = A + B f R ′ ( 0 ) = A − B . f_R(0) \; = \; A + B \qquad \qquad f_R'(0) \; = \; A - B \;. But f R ( 0 ) = ∫ 0 R 1 1 + x 4 d x f R ′ ( 0 ) = − ∫ 0 R x 2 1 + x 4 d x , f_R(0) \; = \; \int_0^R \frac{1}{1+x^4}\,dx \qquad \qquad f_R'(0) \; = \; -\int_0^R \frac{x^2}{1+x^4}\,dx \;, so that f R ( p ) = 1 2 e p ∫ 0 R 1 − x 2 1 + x 4 d x + 1 2 e − p ∫ 0 R 1 + x 2 1 + x 4 d x − ∫ 0 p F R ( u ) sinh ⁡ ( p − u ) d u f_R(p) \; = \; \tfrac12e^p\int_0^R \frac{1-x^2}{1+x^4}\,dx + \tfrac12e^{-p}\int_0^R \frac{1+x^2}{1+x^4}\,dx - \int_0^p F_R(u) \sinh(p-u)\,du . Now F R ( p ) = ∫ 0 R [ cos ⁡ ( p x 2 ) − sin ⁡ ( p x 2 ) ] d x = 1 p ∫ 0 R p [ cos ⁡ ( x 2 ) − sin ⁡ ( x 2 ) ] d x = π 2 p [ C 1 ( R p ) − S 1 ( R p ) ] \begin{array}{rcl} F_R(p) & =& \displaystyle \int_0^R \big[\cos(px^2) - \sin(px^2)\big]\,dx \; =\; \frac{1}{\sqrt{p}} \int_0^{R\sqrt{p}} \big[\cos(x^2) - \sin(x^2)\big]\,dx \\ & = & \sqrt{\frac{\pi}{2p}}\big[C_1(R\sqrt{p}) - S_1(R\sqrt{p})\big] \end{array} and so p F R ( p ) \sqrt{p}F_R(p) is a uniformly bounded function of p p and R R which tends to 0 0 as R → ∞ R \to \infty . Since sinh ⁡ ( p − u ) u \frac{\sinh(p-u)}{\sqrt{u}} is integrable on ( 0 , p ) (0,p) for any p > 0 p > 0 , the Dominated Convergence Theorem gives us that lim ⁡ R → ∞ ∫ 0 p F R ( u ) sinh ⁡ ( p − u ) d u = 0 \lim_{R\to\infty} \int_0^p F_R(u) \sinh(p-u)\,du \; = \; 0 for any p > 0 p > 0 . Since ∫ 0 ∞ 1 1 + x 4 d x = ∫ 0 ∞ x 2 1 + x 4 d x = π 2 2 \int_0^\infty \frac{1}{1+x^4}\,dx \; = \; \int_0^\infty \frac{x^2}{1+x^4}\,dx \; = \; \frac{\pi}{2\sqrt{2}} we deduce finally(!) that lim ⁡ R → ∞ f R ( p ) = π 2 2 e − p , \lim_{R \to \infty} f_R(p) \; = \; \frac{\pi}{2\sqrt{2}}e^{-p} \;, as required.

Mark Hennings - 5 years, 4 months ago

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@Mark Hennings – Wow! That's an awesome method. I'm going to use that henceforth. :)

Aditya Kumar - 5 years, 4 months ago

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@Aditya Kumar – It is a hard work method! Only use it if you have to. Learn what conditions are sufficient to differentiate the integral sign; only get creative if differentiation under the integral sign is not obviously true.

As I think I have said before, learn some Complex Analysis. I think we can agree that my method was quite simple after all!

Mark Hennings - 5 years, 4 months ago

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@Mark Hennings – I'm not so good at complex analysis. I want to learn it. Could you provide me a good pdf for complex analysis? And yes in your solutions which involve complex analysis there isn't any ambiguity.

Aditya Kumar - 5 years, 4 months ago

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@Aditya Kumar – You might try Copson's "An Introduction to the Theory of the Functions of a Complex Variable", available online. It is a bit old-school, but covers the basics. Conway's "Functions of One Complex Variable" is also good.

Mark Hennings - 5 years, 4 months ago

@Mark Hennings – Sir I've provided the proof for fresnel integrals here .

I hope my method is correct.

Aditya Kumar - 5 years, 4 months ago

@Seong Ro have a look at my solution :)

Aditya Kumar - 5 years, 4 months ago

Nice problem! Keep posting more awesome questions :)

Harsh Shrivastava - 5 years, 4 months ago

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