Numerical integration of sharply peaking functions

[says]

Homework Statement

∫ e^{1000((sinx)/x)} dx [0 to 1000 : bound of integration]. Solve this integral of a sharply peaked function without a calculator.

Homework Equations

I'm doing this in relation to statistical thermodynamics - I think I need to use Sterling's Approximation or a gamma function. I'm not really sure how to go about it though. I've had a look at a few definitions of both and they seem a bit too rigorous to grasp yet.

The Attempt at a Solution

 

[stevendaryl]

I can't tell from your problem description what method your instructor wants you to use. However, one way is to write $\frac{1000 sin(x)}{x}$ as a power series:

$\frac{1000 sin(x)}{x} = A + Bx + Cx^2 + ...$

You can then approximate the function by just keeping those three terms (figure out what $A$, $B$ and $C$ are). Then the problem reduces to figuring out $\int_0^{1000} e^{Cx^2} dx$ ($C$ will turn out to be negative). Is that approximately equal to some known integral?

 

[blue_leaf77]

I can't tell from your problem description what method your instructor wants you to use. However, one way is to write $\frac{1000 sin(x)}{x}$ as a power series:

$\frac{1000 \sin(x)}{x} = A + Bx + Cx^2 + ...$

You can then approximate the function by just keeping those three terms (figure out what $A$, $B$ and $C$ are). Then the problem reduces to figuring out $\int_0^{1000} e^{Cx^2} dx$ ($C$ will turn out to be negative). Is that approximately equal to some known integral?

Actually the function $\exp(1000\frac{\sin x}{x})$ approaches unity as x goes to $\pm \infty$ which means its integral over all space is divergent. On the other hand the integral over all space of the approximated function as suggested above is convergent.

 

[stevendaryl]

Actually the function $\exp(1000\frac{\sin x}{x})$ approaches unity as x goes to $\pm \infty$ which means its integral over all space is divergent. On the other hand the integral over all space of the approximated function as suggested above is convergent.

You're right. That doesn't work. The approach I was suggesting only works if the integrand rapidly goes to zero, in which case, you can approximate it by looking at its behavior near $x=0$ (or some other stationary point).

The integral looks something like this:

• The function starts out absolutely huge: $e^{1000}$.
• It drops to almost zero at $x = \frac{3\pi}{2}$
• It goes back up to another huge number: $e^{1000 \frac{2}{5\pi}}$
• It goes back to almost zero at $x = \frac{7\pi}{2}$
• It continues going up and down, with the ups getting smaller and smaller and the downs getting larger and larger, until near $x=1000$, the function is just meandering up and down between the values $e^0$ and $e^1$.

The approximation that I gave is roughly equal to the first region (colored in gray), up to $x= \frac{3\pi}{2}$. But it ignores all the rest. However, we can get an estimate for all the rest, in that it's certainly less than $1000 e^{1000 \frac{2}{5\pi}}$. So if the area of the first region turns out to be really huge compared to this number, then I think my approximation will be close.

 

[says]

I can't tell from your problem description what method your instructor wants you to use. However, one way is to write $\frac{1000 sin(x)}{x}$ as a power series:

$\frac{1000 sin(x)}{x} = A + Bx + Cx^2 + ...$

You can then approximate the function by just keeping those three terms (figure out what $A$, $B$ and $C$ are). Then the problem reduces to figuring out $\int_0^{1000} e^{Cx^2} dx$ ($C$ will turn out to be negative). Is that approximately equal to some known integral?

i know that $\int_{-∞}^{∞} e^{-Cx^2} dx = √(π/C)$ but I'm not really sure where to go from there...

 

[says]

If I represent $1000sin(x)/x$ as a taylor series it equals
$1000 - (500x^2/3) + (25x^4/3)$

$∫e^(1000sin(x)/x) = ∫e^(1000-500x^2/3+25x^4/3)$

$∫ exp(1000) dx + ∫ exp(-500x^2/3) dx + ∫ exp(25x^4/3) dx$

 

[says]

Ok, so I understand the problem a bit more now.

The integral is a sharply peaking function that decays rapidly. All that contributes to the result is the area under the function close to the origin. Instead of doing an expansion of the sharp function, I'm supposed to do one of it's smaller varying logarithmic function. Once I've done that I can expand that function, put it back in the integral and get the result...

 

[says]

$ln (exp(1000(sin(x)/x))$
$= 1000(sin(x)/x)$

series expansion of sin(x) = $x - x^3/3! + x^5/5! ... + ((-1)^n(x)^(2n+1) )/ (2n+1)!$

divide by x to obtain series expansion of sin(x) / x

$sin(x) / x = 1 - x^2/3! + x^4/5! - x^6/7!$

I think that is the correct expansion - from here on I'm not so sure. I put the expansion back into the integral

$∫ ln(exp(1000(sinx/x)) dx = ∫ 1000(sinx/x) dx = 1000 ∫ 1 - x^2/3! + x^4/5! - x^6/7! dx$

[bounds of integration are 0 to ln(1000)] not sure if I apply the log to the bound as well...

$= 1000 [ x - x^3/18 + x^5/600 - x^7/35280 ]$

$= 1000 [ ln(1000) - ln(1000)^3/18 + ln(1000)^5/600 - ln(1000)^7/35280 ] - 0$

$= 1000 [ ln(1000) - ln(1000)^3/18 + ln(1000)^5/600 - ln(1000)^7/35280 ]$

 

[Ray Vickson]

If I represent $1000sin(x)/x$ as a taylor series it equals
$1000 - (500x^2/3) + (25x^4/3)$

$∫e^(1000sin(x)/x) = ∫e^(1000-500x^2/3+25x^4/3)$

$∫ exp(1000) dx + ∫ exp(-500x^2/3) dx + ∫ exp(25x^4/3) dx$

No, $1000 \sin(x)/x$ is not equal to what you wrote. Those are merely the first three terms in the infinite series, and so what you wrote is almost OK for small $x$, say for $0 < x < \pi/2$. The errors get worse and worse as you go to larger $x$, so already by the time you reach $x = \pi$ the errors are perhaps unacceptably large, and they become very bad indeed for still larger $x$.

 

[says]

Ok, so I made anther attempt. As mentioned in post #7, I've taken the natural log of the function. I've then used a taylor series for sin(x)/x

$ln (exp(1000(sin(x)/x))$

$= 1000(sin(x)/x)$

$(sin(x)/x) = ∑ (-1)^n x^{2n} / (2n+1)!$

$1000 ∫ ∑ (-1)^n x^{2n} / (2n+1)!$

$1000 ∑ [ (-1)^n x^{2n+1} / (2n+1) (2n+1)! ]$ | evaluating from 0 to 1000

I'm not sure what I would sum 'n' to though?

 

[Ray Vickson]

Ok, so I made anther attempt. As mentioned in post #7, I've taken the natural log of the function. I've then used a taylor series for sin(x)/x

$ln (exp(1000(sin(x)/x))$

$= 1000(sin(x)/x)$

$(sin(x)/x) = ∑ (-1)^n x^{2n} / (2n+1)!$

$1000 ∫ ∑ (-1)^n x^{2n} / (2n+1)!$

$1000 ∑ [ (-1)^n x^{2n+1} / (2n+1) (2n+1)! ]$ | evaluating from 0 to 1000

I'm not sure what I would sum 'n' to though?

I don't think that expanding it out to a polynomial is helpful, because if quadratic or higher powers of $x$ are included, you cannot write down an exact value anyway, I think you need to go numerical all the way, right from the start.

If I were doing it I would write the integral as $e^{1000} \int_0^{1000} \exp \left( 1000 \frac{\sin x}{x} - 1000 \right) \, dx,$ because the function $f(x) = \exp(1000(-1+\sin(x)/x))$ starts off at $f(0)=1$ and drops rapidly to zero; it is about $10^{-434}$ by the time you get to $x = \pi.$ It remains at or below that level for all $x > \pi.$

 

[says]

I've been told to do an expansion of the function's natural logarithm though, and then put that back into the integral. I understand what you mean in regards to higher orders of x though

 

[says]

I don't understand why when I put the integral into wolfram I get an answer of 1.35*10⁴³³, however when I put the integral into my calculator I get an answer of 1.14*10⁹.