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Verify integration of a sine function

  • Thread starter yungman
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I use part of HS-Scientist derivation in another post thanks to his detail derivation. I want to solving ## \int_0^{\pi} \sin(x sin(\theta)) d\theta##


[tex] \int_0^{\pi} \sin^m(\theta) d\theta=-\frac{1}{m}\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}+\frac{m-1}{m} \int sin^{m-2}(\theta) d\theta= \frac{m-1}{m} \int sin^{m-2}(\theta) d\theta[/tex]
[tex]\hbox{As }\;\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}=0[/tex]


Let ##m=2n+1##
[tex]\Rightarrow\; \int_0^{\pi} \sin^m(\theta) d\theta=\int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2n}{2n+1}\int_0^{\pi} \sin^{2n-1}\theta d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta[/tex]
[tex]\Rightarrow\; \int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta =2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}[/tex]
[tex]\sin x =\sum_0^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)!}[/tex]


[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta=\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!} \int_0^{\pi}\sin^{2n+1}\theta d\theta\;=\;\left[\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!}\right]\left[2\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\right][/tex]

[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{(2n+1)!} \frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))} \;=\;\;\sum_0^{\infty} \frac {(-1)^n 2x^{2n+1}}{3^2 \cdot 5^2\cdot 7^2\cdot \cdot \cdot \cdot (2n+1)^2}[/tex]


I want to verify I am correct in my derivation.

Thanks
 
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Answers and Replies

  • #2
Ray Vickson
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I use part of HS-Scientist derivation in another post thanks to his detail derivation. I want to solving ## \int_0^{\pi} \sin(x sin(\theta)) d\theta##


[tex] \int_0^{\pi} \sin^m(\theta) d\theta=-\frac{1}{m}\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}+\frac{m-1}{m} \int sin^{m-2}(\theta) d\theta= \frac{m-1}{m} \int sin^{m-2}(\theta) d\theta[/tex]
[tex]\hbox{As }\;\left[sin^{m-1}(x)cos(\theta)\right]_0^{\pi}=0[/tex]


Let ##m=2n+1##
[tex]\Rightarrow\; \int_0^{\pi} \sin^m(\theta) d\theta=\int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2n}{2n+1}\int_0^{\pi} \sin^{2n-1}\theta d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta[/tex]
[tex]\Rightarrow\; \int_0^{\pi} \sin^{2n+1}(\theta) d\theta=\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\int_0^{\pi}\sin\theta d\theta =\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\pi[/tex]
[tex]\sin x =\sum_0^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)!}[/tex]


[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta=\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!} \int_0^{\pi}\sin^{2n+1}\theta d\theta\;=\;\left[\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!}\right]\left[\frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\pi\right][/tex]

[tex]\Rightarrow\;\int_0^\pi \sin(x sin(\theta)) d\theta \;=\;\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)!} \frac{2\cdot 4\cdot 6\cdot \cdot \cdot \cdot (2n)}{3\cdot 5\cdot 7\cdot \cdot \cdot (2n+1))}\pi \;=\;\;\sum_0^{\infty} \frac {(-1)^n x^{2n+1}}{2^2 \cdot 4^2\cdot 6^2\cdot \cdot \cdot \cdot (2n)^2}\pi[/tex]


I want to verify I am correct in my derivation.

Thanks
Here is what I get using Maple 11:
f:=sin(x*sin(t));
f := sin(x sin(t)) <--- echo of input
J:=int(f,t =0..Pi);
J := Pi StruveH(0, x) <--- the answer

The function 'StruveH' is non-elementary and is related to solutions of non-homogeneous Bessel differential equations.

We can look at the series expansion:

series(J,x=0,10); <--- get the first 10 terms of the series expn about x = 0:

J = 2*x-(2/9)*x^3+(2/225)*x^5-(2/11025)*x^7+(2/893025)*x^9+O(x^10)

Note that the terms do NOT contain π, so this does not match your solution.
 
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  • #3
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Thanks for the reply. I double checked, I made two mistakes.

1) my in cut and paste in the very last step was wrong. I since corrected the original post already.

2)My other mistake was
[tex]\int_0^{\pi}\sin\theta d\theta=-\cos\theta|_0^{\pi} =-[\cos(\pi)-cos(0)] =-[-1-1] =2\;\hbox {not }\;\pi[/tex]


I did the substitution. I got

[tex]n=0\;\Rightarrow \;\frac {2 x}{1^2}=2 x[/tex]
[tex]n=1\;\Rightarrow \;-\frac {2 x^3}{3^2}=-\frac {2 x^3}{9}[/tex]
[tex]n=2\;\Rightarrow \;\frac {2 x^5}{3^2\cdot 5^2}=\frac {2 x^5}{225}[/tex]
[tex]n=3\;\Rightarrow \;-\frac {2 x^7}{3^2\cdot 5^2\cdot 7^2}=-\frac {2 x^7}{11025}[/tex]
[tex]n=4\;\Rightarrow \;\frac {2 x^9}{3^2\cdot 5^2\cdot 7^2\cdot 9^2}=\frac {2 x^9} {893925}[/tex]

Now it fits your result.

Thanks
 
Last edited:

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