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:

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TopNewestI 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. – Otto Bretscher · 2 years ago

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– Abhineet Nayyar · 2 years 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??Log in to reply

link under "generalization". – Otto Bretscher · 2 years ago

Sure you can use the gamma function with the substitution \(t=x^2\)... just look in Pi Han Goh'sLog in to reply

– Abhineet Nayyar · 2 years ago

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

Just looking at the graph, you know that the answer has to be positive. I will look at your work later... – Otto Bretscher · 2 years ago

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@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? – Abhineet Nayyar · 2 years ago

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– Pi Han Goh · 2 years 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.Log in to reply

– Otto Bretscher · 2 years ago

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

@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...? – Abhineet Nayyar · 2 years ago

I get thatLog in to reply

– Pi Han Goh · 2 years ago

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

– Abhineet Nayyar · 2 years ago

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

– Pi Han Goh · 2 years ago

Something like that. Yeah.Log in to reply

– Abhineet Nayyar · 2 years ago

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

Relevant. – Pi Han Goh · 2 years ago

Nope.Log in to reply

– Abhineet Nayyar · 2 years ago

Well, yayy! i guess...?:pLog in to reply