Mathieu Functions and Hill’s Equation - 28.10 Integral Equations

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DLMF Formula Constraints Maple Mathematica Symbolic
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28.10.E1 2 π ⁢ ∫ 0 π / 2 cos ⁡ ( 2 ⁢ h ⁢ cos ⁡ z ⁢ cos ⁡ t ) ⁢ ce 2 ⁢ n ⁡ ( t , h 2 ) ⁢ d t = A 0 2 ⁢ n ⁢ ( h 2 ) ce 2 ⁢ n ⁡ ( 1 2 ⁢ π , h 2 ) ⁢ ce 2 ⁢ n ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 2 ℎ 𝑧 𝑡 Mathieu-ce 2 𝑛 𝑡 superscript ℎ 2 𝑡 superscript subscript 𝐴 0 2 𝑛 superscript ℎ 2 Mathieu-ce 2 𝑛 1 2 𝜋 superscript ℎ 2 Mathieu-ce 2 𝑛 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos\left(2h% \cos z\cos t\right)\mathrm{ce}_{2n}\left(t,h^{2}\right)\mathrm{d}t=\frac{A_{0}% ^{2n}(h^{2})}{\mathrm{ce}_{2n}\left(\frac{1}{2}\pi,h^{2}\right)}\mathrm{ce}_{2% n}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos@{2h\cos@@{z}\cos@@{t}}\Mathieuce{2n}@{t}{h^{2}}\diff{t} = \frac{A_{0}^{2n}(h^{2})}{\Mathieuce{2n}@{\frac{1}{2}\pi}{h^{2}}}\Mathieuce{2n}@{z}{h^{2}}		 

(2)/(Pi)*int(cos(2*h*cos(z)*cos(t))*MathieuCE(2*n, (h)^(2), t), t = 0..(Pi)/(2)) = ((A[0])^(2*n)((h)^(2)))/(MathieuCE(2*n, (h)^(2), (1)/(2)*Pi))*MathieuCE(2*n, (h)^(2), z)

Divide[2,Pi]*Integrate[Cos[2*h*Cos[z]*Cos[t]]*MathieuC[2*n, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[A, 0])^(2*n)[(h)^(2)],MathieuC[2*n, (h)^(2), Divide[1,2]*Pi]]*MathieuC[2*n, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.10.E2 2 π ⁢ ∫ 0 π / 2 cosh ⁡ ( 2 ⁢ h ⁢ sin ⁡ z ⁢ sin ⁡ t ) ⁢ ce 2 ⁢ n ⁡ ( t , h 2 ) ⁢ d t = A 0 2 ⁢ n ⁢ ( h 2 ) ce 2 ⁢ n ⁡ ( 0 , h 2 ) ⁢ ce 2 ⁢ n ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 2 ℎ 𝑧 𝑡 Mathieu-ce 2 𝑛 𝑡 superscript ℎ 2 𝑡 superscript subscript 𝐴 0 2 𝑛 superscript ℎ 2 Mathieu-ce 2 𝑛 0 superscript ℎ 2 Mathieu-ce 2 𝑛 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cosh\left(2% h\sin z\sin t\right)\mathrm{ce}_{2n}\left(t,h^{2}\right)\mathrm{d}t=\frac{A_{0% }^{2n}(h^{2})}{\mathrm{ce}_{2n}\left(0,h^{2}\right)}\mathrm{ce}_{2n}\left(z,h^% {2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cosh@{2h\sin@@{z}\sin@@{t}}\Mathieuce{2n}@{t}{h^{2}}\diff{t} = \frac{A_{0}^{2n}(h^{2})}{\Mathieuce{2n}@{0}{h^{2}}}\Mathieuce{2n}@{z}{h^{2}}		 

(2)/(Pi)*int(cosh(2*h*sin(z)*sin(t))*MathieuCE(2*n, (h)^(2), t), t = 0..(Pi)/(2)) = ((A[0])^(2*n)((h)^(2)))/(MathieuCE(2*n, (h)^(2), 0))*MathieuCE(2*n, (h)^(2), z)

Divide[2,Pi]*Integrate[Cosh[2*h*Sin[z]*Sin[t]]*MathieuC[2*n, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[A, 0])^(2*n)[(h)^(2)],MathieuC[2*n, (h)^(2), 0]]*MathieuC[2*n, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.10.E3 2 π ⁢ ∫ 0 π / 2 sin ⁡ ( 2 ⁢ h ⁢ cos ⁡ z ⁢ cos ⁡ t ) ⁢ ce 2 ⁢ n + 1 ⁡ ( t , h 2 ) ⁢ d t = - h ⁢ A 1 2 ⁢ n + 1 ⁢ ( h 2 ) ce 2 ⁢ n + 1 ′ ⁡ ( 1 2 ⁢ π , h 2 ) ⁢ ce 2 ⁢ n + 1 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 2 ℎ 𝑧 𝑡 Mathieu-ce 2 𝑛 1 𝑡 superscript ℎ 2 𝑡 ℎ superscript subscript 𝐴 1 2 𝑛 1 superscript ℎ 2 diffop Mathieu-ce 2 𝑛 1 1 1 2 𝜋 superscript ℎ 2 Mathieu-ce 2 𝑛 1 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin\left(2h% \cos z\cos t\right)\mathrm{ce}_{2n+1}\left(t,h^{2}\right)\mathrm{d}t=-\frac{hA% _{1}^{2n+1}(h^{2})}{\mathrm{ce}_{2n+1}'\left(\frac{1}{2}\pi,h^{2}\right)}% \mathrm{ce}_{2n+1}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin@{2h\cos@@{z}\cos@@{t}}\Mathieuce{2n+1}@{t}{h^{2}}\diff{t} = -\frac{hA_{1}^{2n+1}(h^{2})}{\Mathieuce{2n+1}'@{\frac{1}{2}\pi}{h^{2}}}\Mathieuce{2n+1}@{z}{h^{2}}		 

(2)/(Pi)*int(sin(2*h*cos(z)*cos(t))*MathieuCE(2*n + 1, (h)^(2), t), t = 0..(Pi)/(2)) = -((hA[1])^(2*n + 1)((h)^(2)))/(subs( temp=(1)/(2)*Pi, diff( MathieuCE(2*n + 1, (h)^(2), temp), temp$(1) ) ))*MathieuCE(2*n + 1, (h)^(2), z)

Divide[2,Pi]*Integrate[Sin[2*h*Cos[z]*Cos[t]]*MathieuC[2*n + 1, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == -Divide[(Subscript[hA, 1])^(2*n + 1)[(h)^(2)],D[MathieuC[2*n + 1, (h)^(2), temp], {temp, 1}]/.temp-> Divide[1,2]*Pi]*MathieuC[2*n + 1, (h)^(2), z]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.10.E4 2 π ⁢ ∫ 0 π / 2 cos ⁡ z ⁢ cos ⁡ t ⁢ cosh ⁡ ( 2 ⁢ h ⁢ sin ⁡ z ⁢ sin ⁡ t ) ⁢ ce 2 ⁢ n + 1 ⁡ ( t , h 2 ) ⁢ d t = A 1 2 ⁢ n + 1 ⁢ ( h 2 ) 2 ⁢ ce 2 ⁢ n + 1 ⁡ ( 0 , h 2 ) ⁢ ce 2 ⁢ n + 1 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 𝑧 𝑡 2 ℎ 𝑧 𝑡 Mathieu-ce 2 𝑛 1 𝑡 superscript ℎ 2 𝑡 superscript subscript 𝐴 1 2 𝑛 1 superscript ℎ 2 2 Mathieu-ce 2 𝑛 1 0 superscript ℎ 2 Mathieu-ce 2 𝑛 1 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos z\cos t% \cosh\left(2h\sin z\sin t\right)\mathrm{ce}_{2n+1}\left(t,h^{2}\right)\mathrm{% d}t=\frac{A_{1}^{2n+1}(h^{2})}{2\mathrm{ce}_{2n+1}\left(0,h^{2}\right)}\mathrm% {ce}_{2n+1}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos@@{z}\cos@@{t}\cosh@{2h\sin@@{z}\sin@@{t}}\Mathieuce{2n+1}@{t}{h^{2}}\diff{t} = \frac{A_{1}^{2n+1}(h^{2})}{2\Mathieuce{2n+1}@{0}{h^{2}}}\Mathieuce{2n+1}@{z}{h^{2}}		 

(2)/(Pi)*int(cos(z)*cos(t)*cosh(2*h*sin(z)*sin(t))*MathieuCE(2*n + 1, (h)^(2), t), t = 0..(Pi)/(2)) = ((A[1])^(2*n + 1)((h)^(2)))/(2*MathieuCE(2*n + 1, (h)^(2), 0))*MathieuCE(2*n + 1, (h)^(2), z)

Divide[2,Pi]*Integrate[Cos[z]*Cos[t]*Cosh[2*h*Sin[z]*Sin[t]]*MathieuC[2*n + 1, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[A, 1])^(2*n + 1)[(h)^(2)],2*MathieuC[2*n + 1, (h)^(2), 0]]*MathieuC[2*n + 1, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.10.E5 2 π ⁢ ∫ 0 π / 2 sinh ⁡ ( 2 ⁢ h ⁢ sin ⁡ z ⁢ sin ⁡ t ) ⁢ se 2 ⁢ n + 1 ⁡ ( t , h 2 ) ⁢ d t = h ⁢ B 1 2 ⁢ n + 1 ⁢ ( h 2 ) se 2 ⁢ n + 1 ′ ⁡ ( 0 , h 2 ) ⁢ se 2 ⁢ n + 1 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 2 ℎ 𝑧 𝑡 Mathieu-se 2 𝑛 1 𝑡 superscript ℎ 2 𝑡 ℎ superscript subscript 𝐵 1 2 𝑛 1 superscript ℎ 2 diffop Mathieu-se 2 𝑛 1 1 0 superscript ℎ 2 Mathieu-se 2 𝑛 1 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sinh\left(2% h\sin z\sin t\right)\mathrm{se}_{2n+1}\left(t,h^{2}\right)\mathrm{d}t=\frac{hB% _{1}^{2n+1}(h^{2})}{\mathrm{se}_{2n+1}'\left(0,h^{2}\right)}\mathrm{se}_{2n+1}% \left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sinh@{2h\sin@@{z}\sin@@{t}}\Mathieuse{2n+1}@{t}{h^{2}}\diff{t} = \frac{hB_{1}^{2n+1}(h^{2})}{\Mathieuse{2n+1}'@{0}{h^{2}}}\Mathieuse{2n+1}@{z}{h^{2}}		 

(2)/(Pi)*int(sinh(2*h*sin(z)*sin(t))*MathieuSE(2*n + 1, (h)^(2), t), t = 0..(Pi)/(2)) = ((hB[1])^(2*n + 1)((h)^(2)))/(subs( temp=0, diff( MathieuSE(2*n + 1, (h)^(2), temp), temp$(1) ) ))*MathieuSE(2*n + 1, (h)^(2), z)

Divide[2,Pi]*Integrate[Sinh[2*h*Sin[z]*Sin[t]]*MathieuS[2*n + 1, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[hB, 1])^(2*n + 1)[(h)^(2)],D[MathieuS[2*n + 1, (h)^(2), temp], {temp, 1}]/.temp-> 0]*MathieuS[2*n + 1, (h)^(2), z]
Failure Failure Skipped - Because timed out Skipped - Because timed out
28.10.E6 2 π ⁢ ∫ 0 π / 2 sin ⁡ z ⁢ sin ⁡ t ⁢ cos ⁡ ( 2 ⁢ h ⁢ cos ⁡ z ⁢ cos ⁡ t ) ⁢ se 2 ⁢ n + 1 ⁡ ( t , h 2 ) ⁢ d t = B 1 2 ⁢ n + 1 ⁢ ( h 2 ) 2 ⁢ se 2 ⁢ n + 1 ⁡ ( 1 2 ⁢ π , h 2 ) ⁢ se 2 ⁢ n + 1 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 𝑧 𝑡 2 ℎ 𝑧 𝑡 Mathieu-se 2 𝑛 1 𝑡 superscript ℎ 2 𝑡 superscript subscript 𝐵 1 2 𝑛 1 superscript ℎ 2 2 Mathieu-se 2 𝑛 1 1 2 𝜋 superscript ℎ 2 Mathieu-se 2 𝑛 1 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin z\sin t% \cos\left(2h\cos z\cos t\right)\mathrm{se}_{2n+1}\left(t,h^{2}\right)\mathrm{d% }t=\frac{B_{1}^{2n+1}(h^{2})}{2\mathrm{se}_{2n+1}\left(\frac{1}{2}\pi,h^{2}% \right)}\mathrm{se}_{2n+1}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin@@{z}\sin@@{t}\cos@{2h\cos@@{z}\cos@@{t}}\Mathieuse{2n+1}@{t}{h^{2}}\diff{t} = \frac{B_{1}^{2n+1}(h^{2})}{2\Mathieuse{2n+1}@{\frac{1}{2}\pi}{h^{2}}}\Mathieuse{2n+1}@{z}{h^{2}}		 

(2)/(Pi)*int(sin(z)*sin(t)*cos(2*h*cos(z)*cos(t))*MathieuSE(2*n + 1, (h)^(2), t), t = 0..(Pi)/(2)) = ((B[1])^(2*n + 1)((h)^(2)))/(2*MathieuSE(2*n + 1, (h)^(2), (1)/(2)*Pi))*MathieuSE(2*n + 1, (h)^(2), z)

Divide[2,Pi]*Integrate[Sin[z]*Sin[t]*Cos[2*h*Cos[z]*Cos[t]]*MathieuS[2*n + 1, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[B, 1])^(2*n + 1)[(h)^(2)],2*MathieuS[2*n + 1, (h)^(2), Divide[1,2]*Pi]]*MathieuS[2*n + 1, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.10.E7 2 π ⁢ ∫ 0 π / 2 sin ⁡ z ⁢ sin ⁡ t ⁢ sin ⁡ ( 2 ⁢ h ⁢ cos ⁡ z ⁢ cos ⁡ t ) ⁢ se 2 ⁢ n + 2 ⁡ ( t , h 2 ) ⁢ d t = - h ⁢ B 2 2 ⁢ n + 2 ⁢ ( h 2 ) 2 ⁢ se 2 ⁢ n + 2 ′ ⁡ ( 1 2 ⁢ π , h 2 ) ⁢ se 2 ⁢ n + 2 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 𝑧 𝑡 2 ℎ 𝑧 𝑡 Mathieu-se 2 𝑛 2 𝑡 superscript ℎ 2 𝑡 ℎ superscript subscript 𝐵 2 2 𝑛 2 superscript ℎ 2 2 diffop Mathieu-se 2 𝑛 2 1 1 2 𝜋 superscript ℎ 2 Mathieu-se 2 𝑛 2 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin z\sin t% \sin\left(2h\cos z\cos t\right)\mathrm{se}_{2n+2}\left(t,h^{2}\right)\mathrm{d% }t=-\frac{hB_{2}^{2n+2}(h^{2})}{2\mathrm{se}_{2n+2}'\left(\frac{1}{2}\pi,h^{2}% \right)}\mathrm{se}_{2n+2}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\sin@@{z}\sin@@{t}\sin@{2h\cos@@{z}\cos@@{t}}\Mathieuse{2n+2}@{t}{h^{2}}\diff{t} = -\frac{hB_{2}^{2n+2}(h^{2})}{2\Mathieuse{2n+2}'@{\frac{1}{2}\pi}{h^{2}}}\Mathieuse{2n+2}@{z}{h^{2}}		 

(2)/(Pi)*int(sin(z)*sin(t)*sin(2*h*cos(z)*cos(t))*MathieuSE(2*n + 2, (h)^(2), t), t = 0..(Pi)/(2)) = -((hB[2])^(2*n + 2)((h)^(2)))/(2*subs( temp=(1)/(2)*Pi, diff( MathieuSE(2*n + 2, (h)^(2), temp), temp$(1) ) ))*MathieuSE(2*n + 2, (h)^(2), z)

Divide[2,Pi]*Integrate[Sin[z]*Sin[t]*Sin[2*h*Cos[z]*Cos[t]]*MathieuS[2*n + 2, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == -Divide[(Subscript[hB, 2])^(2*n + 2)[(h)^(2)],2*(D[MathieuS[2*n + 2, (h)^(2), temp], {temp, 1}]/.temp-> Divide[1,2]*Pi)]*MathieuS[2*n + 2, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
28.10.E8 2 π ⁢ ∫ 0 π / 2 cos ⁡ z ⁢ cos ⁡ t ⁢ sinh ⁡ ( 2 ⁢ h ⁢ sin ⁡ z ⁢ sin ⁡ t ) ⁢ se 2 ⁢ n + 2 ⁡ ( t , h 2 ) ⁢ d t = h ⁢ B 2 2 ⁢ n + 2 ⁢ ( h 2 ) 2 ⁢ se 2 ⁢ n + 2 ′ ⁡ ( 0 , h 2 ) ⁢ se 2 ⁢ n + 2 ⁡ ( z , h 2 ) 2 𝜋 superscript subscript 0 𝜋 2 𝑧 𝑡 2 ℎ 𝑧 𝑡 Mathieu-se 2 𝑛 2 𝑡 superscript ℎ 2 𝑡 ℎ superscript subscript 𝐵 2 2 𝑛 2 superscript ℎ 2 2 diffop Mathieu-se 2 𝑛 2 1 0 superscript ℎ 2 Mathieu-se 2 𝑛 2 𝑧 superscript ℎ 2 {\displaystyle{\displaystyle\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos z\cos t% \sinh\left(2h\sin z\sin t\right)\mathrm{se}_{2n+2}\left(t,h^{2}\right)\mathrm{% d}t=\frac{hB_{2}^{2n+2}(h^{2})}{2\mathrm{se}_{2n+2}'\left(0,h^{2}\right)}% \mathrm{se}_{2n+2}\left(z,h^{2}\right)}}
\frac{2}{\pi}\int_{0}^{\ifrac{\pi}{2}}\cos@@{z}\cos@@{t}\sinh@{2h\sin@@{z}\sin@@{t}}\Mathieuse{2n+2}@{t}{h^{2}}\diff{t} = \frac{hB_{2}^{2n+2}(h^{2})}{2\Mathieuse{2n+2}'@{0}{h^{2}}}\Mathieuse{2n+2}@{z}{h^{2}}		 

(2)/(Pi)*int(cos(z)*cos(t)*sinh(2*h*sin(z)*sin(t))*MathieuSE(2*n + 2, (h)^(2), t), t = 0..(Pi)/(2)) = ((hB[2])^(2*n + 2)((h)^(2)))/(2*subs( temp=0, diff( MathieuSE(2*n + 2, (h)^(2), temp), temp$(1) ) ))*MathieuSE(2*n + 2, (h)^(2), z)

Divide[2,Pi]*Integrate[Cos[z]*Cos[t]*Sinh[2*h*Sin[z]*Sin[t]]*MathieuS[2*n + 2, (h)^(2), t], {t, 0, Divide[Pi,2]}, GenerateConditions->None] == Divide[(Subscript[hB, 2])^(2*n + 2)[(h)^(2)],2*(D[MathieuS[2*n + 2, (h)^(2), temp], {temp, 1}]/.temp-> 0)]*MathieuS[2*n + 2, (h)^(2), z]
Failure Aborted Skipped - Because timed out Skipped - Because timed out
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