java - Writing a formula for an infinite sum, why doesn't this work as written? -

i writing infinite sum inside java match -

sqrt(t)*sech(t)^2 dt t=0 t=infinity (an infinite sum starting t = 0 , ending @ t = infinity. referencing wolfram alpha (mathematica) compare results).

in more mathematical terms, (essentially) program doing. note squaring (hyperbolic) secant. although, maximum infinity-

integrate sqrt(t)*sech(t)^2 dt t=0 t=1000 

to match infinite sum, wrote short program below.

public class testsum {    public static void main(string[] args) {            newform(0.5);    }     public static double newform(double s) {     int n = 0;     double currentsum = 0;      while (n < 1000) {         double sech = 1 / math.cosh(n);         double squared = math.pow(sech, 2);         currentsum = ((math.pow(n, s))*squared) + currentsum;         if(n == 999)             system.out.println("the current sum " + currentsum);         n++;     }     return currentsum;    } } 

when plug mathematica/wolfram -

integrate sqrt(t)*sech(t)^2 dt t=0 t=1000 integral_0^1000 sqrt(t) sech^2(t) dt = 0.758128 

the result running program -

run: current sum 0.5401365941579325 build successful (total time: 0 seconds) 

i pretty sure mathematica isn't wrong. wrong program?

your solution not accurate enough.

an integral can approximated riemann sum

see riemann sum on wikipedia.

the result gets better smaller delta x (or in case delta t) becomes.

in solution delta t = 1, approximation not good.

the possible solution approximate result better use:

public class testsum {    public static void main(string[] args) {           double result= integrate(0, 1000);           system.out.print("result = " + result );    }     public static double integrate(double start, double end) {     double currentintegralvalue = 0;     double dt=0.01d;     double t = start;      while (math.abs(end - t) >= dt && t-end < 0) {         currentintegralvalue += fn(t)*dt;         t += dt;     }     return currentintegralvalue;    }     private static double fn(double t) {         double sech = 1 / math.cosh(t);         double squared = math.pow(sech, 2);         return  ((math.pow(t, 0.5))*squared);     } } 

result = 0.7579201343666041

you can further improve result using smaller dt.

dt=0.00001d 

result = 0.7581278135568323