# 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 : $\int _{ 0 }^{ \infty }{ \sin { ({ x }^{ 2 }) } dx }$
And I have solved it like this:

Note by A Former Brilliant Member
5 years, 10 months ago

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

- 5 years, 10 months ago

@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?

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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...?

- 5 years, 10 months ago

You can interpret them as the difference between the nth positive area and nth negative area converge to 0.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Something like that. Yeah.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Nope. Relevant.

- 5 years, 10 months ago

Well, yayy! i guess...?:p

- 5 years, 10 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.

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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??

- 5 years, 10 months ago

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

- 5 years, 10 months ago

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

- 5 years, 10 months ago