Integration?

Question

How would I go about integrating this function:

sqrt(2/a) integral (0-->a) [sin (n*pi*x / (a))] / (sqrt (x)) dx

This is Integral of sin (n*pi*x/a) / sqrt(x)

It's kind of hard to explain? Sorry. I hope someone can help.

Love,
Mary

Sorry, guys! I set up the problem wrong.. :( I didn't have sqrt of x it's actually sqrt of a which is a constant and be brought out. Eek. Sorry!

Love,
Mary

Answer 1

To make life simpler, let's let a = 1
With an appropriate substitution, we basically reduce
the integral to
∫ sin(x)/ √x dx.
Let u = √x, x = u² dx = 2u du.
So we get
2*∫ sin u² du.
But this is a Fresnel integral and cannot be integrated
in terms of elementary functions.
So your original problem can't be done in terms
of elementary antiderivatives.
Probably you need to use infinite series
or Simpson's rule to evaluate it.

Answer 2

Mary,

I know you've been working quantum chemistry problems, so can I ask, does this arise from a quantum mechanics problem? If so, the sqrt(x) in the denominator is somewhat unusual for the typical introductory quantum mechanics problems, so I wonder if you perhaps made an error setting up the problem. If this is part of a larger problem, maybe it would help to ask a question about that problem directly.

If this isn't a quantum mechanics problem, I apologize.

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Edit: Sometimes people get in trouble evaluating integrals in quantum mechanics in spherical coordinates, because they forget that the 3D spherical volume element is dV = r^2 sin(theta) dr d(theta) d(phi). This extra factor of r^2 helps to cancel some strange denominators. So, if by "x" you really mean a radial distance r, then you might want to re-examine the volume element of integration.

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Edit 2: I take it back, the sin(n*pi*x/a) suggests a problem in Cartesion coordinates, not spherical coordinates. But if this is a quantum mechanics problem, I still suspect that there may be a problem somewhere with how you set it up.

Answer 3

Note: we can re-write want we want to integrate as:

sin(k*x)/sqrt(x) where k = n*pi/a, because they are arbitrary constants

Now let us substitute:

kx = arcsinp

This implies:
sin(kx) = p

k.dx = 1/sqrt(1-p^2).dp

Now change your integration limits. Now your integration turns into this:

INT[p/sqrt(1-p^2)].dp

You may/may not be able to do this straight away, if not, substitute again:
s = 1 - p^2
Then:
ds = -2p.dp
Pull out the -2 factor, change integration limits again, and your integration now becomes:

-2*INT(1/s).ds

And you can do that easy. Remeber to keep changing your integration limits when you substitute, i.e. it will no longer be from 0-->a.

Answer 4

▬ OMG. skip the garbage below till next ▬

♠ y(x)*dx = dx* sin(n*pi*(x/a)) *√(2/(ax));
x=a*t^2, dx=2at*dt;
y(t)*dt = 2at*dt *sin(n*pi*t^2) *√(2/(at)^2) =
= 2√2*dt* sin(n*pi*t^2) for t=0 until 1;
♣ tailor says: sin(x) = x -x^3/3! +x^5/5! -+-+;
y(t)*dt = -2√2*dt* ∑(-1)^j *(n*pi*t^2)^(2j-1) / (2j-1)!,
where integer j ≥ 1; integrating gives:
♦ Y(t) = -2√2 ∑(-1)^j *(n*pi)^(2j-1) * t^(4j-1)) /{(4j-1) (2j-1)!};
Y(1,n) = -2√2 ∑(-1)^j *(n*pi)^(2j-1) /{(4j-1) (2j-1)!}, j=1,2,3,4,,,
♦ hark, it does not depend on a;
♦ put this formula into Excel; it seams first to diverge then rapidly converges, e.g.
Y(1,7)=0.442175;
♦ meanwhile trapezoid method with step 1/1024 gives Y(1,7)=0.442165;

▬ Y=∫ dx* sin(n*pi*(x/a)) = -cos(n*pi*(x/a)) / (n*pi/a)=
= {for x=0 until a} = -((-1)^n -1)/ (n*pi/a);
how to treat it?
1)If n is even Y=0;
2)if n is odd Y=2a/(n*pi);
3)if n=0 Y = const known to you!
▬ start voting yourself or no more help!

Answer 5

Not easily. I typed this in to my symbolic math system and it came back with a Fresnel function. So, unless you're in advanced calculus or optics, I suspect you may be biting off more than you bargained for.

For what it's worth...

2^(1/2)*pi^(1/2)/(n*pi/a)^(1/2) *
FresnelS(2^(1/2)*pi^(1/2)/(n*pi/a)^(1/2)*n*x^(1/2)/a)

Answer 6

At first you should change the variable.
Use
u = √(nπx/a),
du = 1/2 1/√x √(nπ/a) dx,
and rewrite your integral as

Integral [0..√a] of sin(u²) du

As the other guy said this integral is denoted as fresnel integral.

Analytical expression in elementary functions can be drived in case a=infinity, though.
In this case
Integral [0..oo] of sin(u²) du =
Integral [0..oo] of Im(exp(iu²)) du =
Im [Integral [0..oo] of exp(iu²) du] =
Im [1/√i Integral [0..oo] of exp(z²) dz]

The latter is integrable along path 0->(oo + 0i).
It also vanishes along along infinutely remote arc
(oo + 0i) -> oo*√i
Therefore you can find it along the desirable path
0 -> oo*√i

Answer 7

im not familiar with integrating with two different variables within the integral, but i would assume you need either a substitution or use integration by parts