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:

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

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??

Log in to reply

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

Log in to reply

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...

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?

Log in to reply

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

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...?

I get thatLog in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Relevant.

Nope.Log in to reply

Log in to reply