Question

Integral challenge\[\int_{0}^{\infty} \sin (x^2) \ \text{d}x = ?\]

Answer 1

.

Answer 2

u substitution?

Answer 3

like let u = x^2
du = 2dx
\[\frac{du}{2} = dx\]

and then -cos x
because the derivative of cos x is -sin x and if we have that extra negative we can multiply . A negative x negative is positive
-(-sinx) = sinx

Answer 4

sorry I'm typing this while dealing with a stomachache

Answer 5

I just typo'ed dropped the x :(

Answer 6

u = x^2
du = 2x dx
\[\frac{du}{2x} = dx\]

Answer 7

so to gain sin(u) back

I need -cos(u)
because the derivative of cos(u) is -sin(u) and with that extra negative in place... a negative x a negative is a positive so -(-sin(u)) ->sin (u)

\[\frac{-1}{2x}\cos(x^2)+C\]

Answer 8

FORGET IT! It looks innocent, yet when I try to do it, it's all screwed up.

Answer 9

nugh! there has got to be a way to get rid of that second part!!!

Answer 10

first part is fine... second is leave cos(x^2) deal with that fraction which is quotient rule

Answer 11

hint:
we can write this:
\[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\sin } \left( {{x^2}} \right)\;dx\]

Answer 12

now I change variable:
\[\large {x^2} = z\]
where z is the new integration variable

Answer 13

There is no anti-derivative for \(\sin x^2\) in terms of elementary functions. But there is a way to solve this integral with elementary functions. Also a little bit of Laplace transforms needed here and first thing to do is changing it to a double integral.

Answer 14

\[\large \int_0^{ + \infty } {\sin \left( {{x^2}} \right)} \;dx = \int_0^{ + \infty } {\sin \left( z \right)} \;\frac{{dz}}{{2\sqrt z }}\]

Answer 15

In case you don't wanna use complex analysis. Michele is on the right track till now.

Answer 16

the subsequent function:
\[\large \frac{{\sin \left( z \right)}}{{2\sqrt z }}\]
is holomorphic in all z-plane except z=0

Answer 17

you wanna use complex analysis?

Answer 18

may I apply complex analysis?

Answer 19

oh yeah, no problem

Answer 20

ok! Then the line of integration is: