This integral solveable [itex]\int^{3.5}_{0.5}(sin(x*π)+1)^adx[/itex]?

[no_alone]

Hello I am trying to find a normalizing factor for
\int^{3.5}_{0.5}(sin(x*π)+1)^1dx

I want that the integral of \int^{3.5}_{0.5}(sin(x*π)+1)^a dx *factor = \int^{3.5}_{0.5}(sin(x*π)+1)^1dx

So for this I need to solve the integral
\int^{3.5}_{0.5}(sin(x*π)+1)^adx

But I did not mange to solve it.
I did found that when I do = a = [1,2,3,4,5,6,7]
I find in wolfram this:
http://www.wolframalpha.com/input/?i=2+3+5+8.75+15.75+28.875

in the end of the page we can see a possible sequenct identification:
α_{n} = \frac{2^{n}* (\frac{3}{2} )_{n-1} }{(2)_{n-1}}

when σ_n is the Pochhammer symbol.

Thank You.

 

[DrClaude]

Here is what Mathematica gives:
$$
\int_{\frac{1}{2}}^{\frac{7}{2}} (\sin (\pi x)+1)^a \, dx =
\frac{2^a \left(3 \sqrt{\pi } \Gamma \left(a+\frac{1}{2}\right)-2 \Gamma (a+1) B_0\left(a+\frac{1}{2},\frac{1}{2}\right)\right)}{\pi \Gamma (a+1)}
$$
provided $\Re(a) > -1/2$. $\Gamma$ is the gamma function and $B_0$ is the incomplete beta function.

 

[D H]

If appears that you are only worried about a being a positive integer. If that's the case, denote $k_a \equiv \int_{1/2}^{7/2} (1+\sin(\pi x))^a\,dx$. Expand this as $k_a = \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\,dx + \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\sin(\pi x)\,dx = k_{a-1} + I_a$ where $I_a \equiv \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\sin(\pi x)\,dx$. Integrate this by parts. You should get a linear combination of $k_{a-1}$ and $k_a$. Solve $k_a = k_{a-1}+I_a$ for $k_a$. This will give a recursive relation for $k_a$ in terms of $k_{a-1}$. You know $k_1$, so this will let you solve for $k_a$ in terms of $a$.

 

[no_alone]

Here is what Mathematica gives:
$$
\int_{\frac{1}{2}}^{\frac{7}{2}} (\sin (\pi x)+1)^a \, dx =
\frac{2^a \left(3 \sqrt{\pi } \Gamma \left(a+\frac{1}{2}\right)-2 \Gamma (a+1) B_0\left(a+\frac{1}{2},\frac{1}{2}\right)\right)}{\pi \Gamma (a+1)}
$$
provided $\Re(a) > -1/2$. $\Gamma$ is the gamma function and $B_0$ is the incomplete beta function.

Thank You.
Do you, but this will not help me, Because in your solution, I need to calculate the gamma function, and this is also an integral. I can do it computationally .. But I can just calculate the first integral.

If appears that you are only worried about a being a positive integer. If that's the case, denote $k_a \equiv \int_{1/2}^{7/2} (1+\sin(\pi x))^a\,dx$. Expand this as $k_a = \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\,dx + \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\sin(\pi x)\,dx = k_{a-1} + I_a$ where $I_a \equiv \int_{1/2}^{7/2} (1+\sin(\pi x))^{a-1}\sin(\pi x)\,dx$. Integrate this by parts. You should get a linear combination of $k_{a-1}$ and $k_a$. Solve $k_a = k_{a-1}+I_a$ for $k_a$. This will give a recursive relation for $k_a$ in terms of $k_{a-1}$. You know $k_1$, so this will let you solve for $k_a$ in terms of $a$.

Thank You D H.. But I also a=0.1 0.5 1.5 ... I am also worried about a being not integral ( but I do not care if it is not positive ).
Your solution will help me for integer a. :-)

 

[JJacquelin]

. I need to calculate the gamma function, and this is also an integral.

So, do you mean that you cannot calculate the exp function, nor the log function ? They are also integals, just as the gamma function and many other special functions :
http://fr.scribd.com/doc/14623310/S...tions-Safari-au-pays-des-fonctions-speciales-

 

[no_alone]

So, do you mean that you cannot calculate the exp function, nor the log function ? They are also integals, just as the gamma function and many other special functions :
http://fr.scribd.com/doc/14623310/S...tions-Safari-au-pays-des-fonctions-speciales-

Thank You, I write a program that should normalize the integral for every α. and I write it in a wrapper for c.. all I have are those functions:
http://www.neuron.yale.edu/neuron/static/docs/nmodl/nmodlfunc.html

I do not have the gamma function and incomplete beta function.

 

[DrClaude]

I do not have the gamma function and incomplete beta function.

You should use GSL: http://www.gnu.org/software/gsl/man...-Beta-Functions.html#Gamma-and-Beta-Functions