Guys!

I'm deriving something that I feel is cool, and I am so close to the answer, but i'm getting both the positive as well as a negative value as an answer. I need some solid proof to reject one of the answers as the internet only gives the positive value as an answer...Help?

The question is : 0sin(x2)dx\int _{ 0 }^{ \infty }{ \sin { ({ x }^{ 2 }) } dx }
And I have solved it like this:

#Integration #Complex

Note by A Former Brilliant Member
5 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Just looking at the graph, you know that the answer has to be positive. I will look at your work later...

Otto Bretscher - 5 years, 8 months ago

Log in to reply

@Otto Bretscher I thought so too, but by looking at the graph, we can say that the first positive area is large...But if the limit continues upto infinity, we can't say that the total sum is positive, can we?

A Former Brilliant Member - 5 years, 8 months ago

Log in to reply

Oh sure we can... each region above the axis is a little larger than the next region below.

Otto Bretscher - 5 years, 8 months ago

Log in to reply

@Otto Bretscher I get that @Otto Bretscher and @Pi Han Goh ...But although the first positive area is more than the first negative area....i can similarly argue, that the next positive area is less than the first negative area....which goes on upto infinity...?

A Former Brilliant Member - 5 years, 8 months ago

Log in to reply

@A Former Brilliant Member You can interpret them as the difference between the nth positive area and nth negative area converge to 0.

Pi Han Goh - 5 years, 8 months ago

Log in to reply

@Pi Han Goh Hmmm, so we take them as consecutive areas, is that what you're saying...?

A Former Brilliant Member - 5 years, 8 months ago

Log in to reply

@A Former Brilliant Member Something like that. Yeah.

Pi Han Goh - 5 years, 8 months ago

Log in to reply

@Pi Han Goh Alright, well thanks...So, there's no problem in my solution, right? It's cool?

A Former Brilliant Member - 5 years, 8 months ago

Log in to reply

@A Former Brilliant Member Nope. Relevant.

Pi Han Goh - 5 years, 8 months ago

Log in to reply

@Pi Han Goh Well, yayy! i guess...?:p

A Former Brilliant Member - 5 years, 8 months ago

Your working is right. It's easy to show why the answer must be positive by looking at the first positive area and you can see that the curve becomes more erratic.

Pi Han Goh - 5 years, 8 months ago

I don't really get what you are doing. Since your tt is an imaginary quantity, what do you mean by 0...dt\int_{0}^{\infty}...dt ? It's better to use the substitution t=x2t=x^2, and things fall into place nicely.

Otto Bretscher - 5 years, 8 months ago

Log in to reply

Hmm with the substitution t=x^2 we cant use gamma theorem...for that we need a term of e^-x where x is the variable...thats why i had to take t as an imaginary quantity, but i dont think complex or real makes any difference, does it??

A Former Brilliant Member - 5 years, 8 months ago

Log in to reply

Sure you can use the gamma function with the substitution t=x2t=x^2... just look in Pi Han Goh's link under "generalization".

Otto Bretscher - 5 years, 8 months ago

Log in to reply

@Otto Bretscher Oh yes...got it...i forgot that i did start with this substitution buy changed the path in between...THANKS!! :)

A Former Brilliant Member - 5 years, 8 months ago
×

Problem Loading...

Note Loading...

Set Loading...