I want to do an integral like the picture. In the end, "x" will remain in the result expression. Anybody can help me?

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Roger Stafford
Roger Stafford 2014년 12월 5일
편집: Roger Stafford 2014년 12월 5일
I doubt that you will be able to find an explicit expression for this integral in terms of x. However, it does have an explicit power series expansion in terms of odd powers of x which converges for all x satisfying abs(x) < 1. Is that of interest to you, Chen? Otherwise I believe you will have to settle for using a numerical integration function in which each value of x must be specified.
Correction: The above power series would actually converge for all x.
Chen
Chen 2014년 12월 8일
actually, this integration is inside a larger integration. through the integration i posted, i want to obtain an expression with the only variation, x.
Roger Stafford
Roger Stafford 2014년 12월 9일
Even though no explicit expression is possible for this integral which depends on x, it is quite possible for you to devise a matlab function which takes x as its input and gives as an output the value of your integral using either numerical integration or the power series I have given you. In turn this function can be used for creating or helping to create the integrand for an outer integral.
See http://www.mathworks.com/help/matlab/ref/integral.html for the requirements on integrand functions for the numerical quadrature function 'integral', including the necessity for them to accept vector inputs.

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Roger Stafford
Roger Stafford 2014년 12월 7일
편집: Roger Stafford 2014년 12월 7일

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Just in case it is of interest to you, here is that infinite series expansion of your integral which I referred to:
-2*pi*((x/2)/(0!*1!) + (x/2)^3/(1!*2!) + (x/2)^5/(2!*3!) + ...
+ (x/2)^(2*n-1)/((n-1)!*n!) + ... )
It is obtained using the Taylor Series expansion of your function about x = 0 and using the identity
int(cos(t)^n,'t',0,pi/2) = (1*3*5*...*(n-1))/(2*4*6*...*n)*pi/2
where n is an even integer.

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