Linked Questions

12
votes
4answers
18k views

Some way to integrate $\sin(x^2)$? [duplicate]

Because the straight forward approach involves Fresnel integrals I thought about a different approach of taking the imaginary part of $\int_{-\infty}^{\infty}\exp{(ix^2)} \, dx $ but have no idea how ...
2
votes
1answer
247 views

How to calculate $\int_{0}^{+\infty} {\sin {x^2}\mathrm{d}x}$? [duplicate]

Possible Duplicate: Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? How to calculate $\displaystyle\int_{0}^{+\infty} {\sin {x^2}\mathrm{d}x}$ ?
3
votes
1answer
164 views

Evaluate $\int_{0}^{\infty} \cos(x^2)dx $ [duplicate]

Prove that the above integral is equal to $\frac{\sqrt{2\pi}}{2}$ I have already tried expanding using $\cos$ identity and also taking Laplace for it. I am getting nowhere with this.
170
votes
19answers
45k views

Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$

How to prove $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
30
votes
6answers
17k views

Prove: $\int_0^\infty \sin (x^2) \, dx$ converges.

$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does. I'm trying to understand why and would like some help in working towards a formal proof.
22
votes
6answers
3k views

Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$

Numerically it seems to be true that $$ \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. $$ Any ideas how to prove this?
33
votes
5answers
2k views

Generalizing the trick for integrating $\int_{-\infty}^\infty e^{-x^2}\mathrm dx$?

There is a well-known trick for integrating $\int_{-\infty}^\infty e^{-x^2}\mathrm dx$, which is to write it as $\sqrt{\int_{-\infty}^\infty e^{-x^2}\mathrm dx\int_{-\infty}^\infty e^{-y^2}\mathrm dy}$...
30
votes
4answers
2k views

tough integral involving $\sin(x^2)$ and $\sinh^2 (x)$

I ran across this integral I get no where with. Can someone suggest a method of attack?. $$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$ I tried series, ...
33
votes
4answers
1k views

Proof only by transformation that : $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

This was a question in our exam and I did not know which change of variables or trick to apply How to show by inspection ( change of variables or whatever trick ) that $$ \int_0^\infty \cos(x^2)...
7
votes
5answers
2k views

Proof of Fresnel's Integral

So, my teacher wants us to prove the Fresnel's integral: $$\int_0^\infty\cos(x^2)dx=\sqrt{\frac{\pi}{8}}$$ The problem is that we cannot use complex analysis to prove that and we should do that ...
9
votes
3answers
765 views

Meaning and example(s) of Qiaochu's quote.

I happen to come across this page http://math.uchicago.edu/~chonoles/quotations.html which contains some beautiful quotes by various mathematicians and I came across Qiaochu's quote as claimed by the ...
9
votes
4answers
2k views

Do integrable functions vanish at infinity? [duplicate]

If $f$ is a real-valued function that is integrable over $\mathbb{R}$, does it imply that $$f(x) \to 0 \text{ as } |x| \to \infty? $$ When I consider, for simplicity, positive function $f$ which is ...
15
votes
2answers
396 views

Evaluating $ \int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$

$$\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$$ This is a problem from the Pi Mu Epsilon Journal, and I'm having great trouble answering it. I've tried some substitutions ...
7
votes
2answers
1k views

Trig Fresnel Integral

$$\int_{0}^{\infty }\sin(x^{2})dx$$ I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you. I have not done complex analysis (only ...
1
vote
6answers
3k views

if the improper integral $\int^\infty_a f(x)\,dx$ converges, then $\lim_{x→∞}f(x)=0$ [closed]

I need to prove that: $$\lim_{x→∞}f(x)=0$$ if $$\displaystyle∫^∞_af(x)\,dx$$ converges. I need a proof or an specific, and if possible simple, counterexample. Would really appreciate your help! ...

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