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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 Abhineet Nayyar
2 years ago

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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. Otto Bretscher · 2 years ago

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@Otto Bretscher 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?? Abhineet Nayyar · 2 years ago

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@Abhineet Nayyar Sure you can use the gamma function with the substitution \(t=x^2\)... just look in Pi Han Goh's link under "generalization". Otto Bretscher · 2 years ago

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@Otto Bretscher Oh yes...got it...i forgot that i did start with this substitution buy changed the path in between...THANKS!! :) Abhineet Nayyar · 2 years 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... Otto Bretscher · 2 years ago

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@Otto Bretscher @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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@Abhineet Nayyar 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 · 2 years ago

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@Abhineet Nayyar Oh sure we can... each region above the axis is a little larger than the next region below. Otto Bretscher · 2 years ago

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@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...? Abhineet Nayyar · 2 years ago

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@Abhineet Nayyar You can interpret them as the difference between the nth positive area and nth negative area converge to 0. Pi Han Goh · 2 years ago

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@Pi Han Goh Hmmm, so we take them as consecutive areas, is that what you're saying...? Abhineet Nayyar · 2 years ago

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@Abhineet Nayyar Something like that. Yeah. Pi Han Goh · 2 years ago

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@Pi Han Goh Alright, well thanks...So, there's no problem in my solution, right? It's cool? Abhineet Nayyar · 2 years ago

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@Abhineet Nayyar Nope. Relevant. Pi Han Goh · 2 years ago

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@Pi Han Goh Well, yayy! i guess...?:p Abhineet Nayyar · 2 years ago

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