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Integrate a power beside a sine or a cosine as incomplete gamma functions - #2085
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…ions A power of x or of a linear that is not a polynomial, beside sin(u)^k cos(u)^j, is integrated through Euler's formula: sin(u)^k cos(u)^j is (2 i)^(-k) 2^(-j) sum C(k, r) (-1)^(k - r) C(j, t) e^(i (2 r - k + 2 t - j) u), each e^(i q u) beside the power is a gamma function of the gamma rule for a power beside an exponential, and the term with q = 0 is the power's own integral. Where any term is declined -- a whole power, whose integral is elementary and answered by parts, or a half-whole order, which is Fresnel's -- so is the integrand. The gamma functions' arguments are imaginary on the real line, off the cut. Rubi's suite on the harness that reads GAMMA, against master 23c8eca: 48,700 -> 48,850 of 49,052, 0 wrong, no solved row lost, no row to a timeout, no answer's allocation grown. The gains are 57 in 4.1.12, 26 in 4.1.10, 25 in 4.2.10, 18 in 4.2.12, 12 in 5.1.4, 9 in 4.7.3, 2 in 8.4 and one Moses problem. Four definite integrals agree with quadrature to 25 digits, real. Part of #2075. Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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Integrates a power of
xor of a linear that is not a polynomial, besidesin(u)^k cos(u)^j, through Euler's formula. Part of #2075.sin(u)^k cos(u)^jis a sum ofe^(i q u). Each term beside the power is a gamma function of the rule from #2083, and the term withq = 0is the power's own integral. Where any term is declined, so is the integrand: a whole power keeps its elementary answer by parts, and a half-whole order, which is Fresnel's, stays unanswered as before. The gamma functions' arguments are imaginary on the real line, off the cut, so the branch question of #2080 does not arise.(c + d x)^m sin(a + b x)x^4 (a + b sin(c + d x^3))^2sin(c + d x^3)xaloneMeasured on Rubi's suite, read with the harness that maps
GAMMA, against master23c8eca1, the whole corpus each side:The gains are 57 in 4.1.12, 26 in 4.1.10, 25 in 4.2.10, 18 in 4.2.12, 12 in 5.1.4, 9 in 4.7.3, 2 in 8.4 and one of Moses's problems. Family 4 goes from 14,107 to 14,242 of 14,262. Time and allocation per stream are within 3%, and no answer's allocation grew.
By value. Four definite integrals agree with quadrature to 25 digits, and each is real:
integral((5/4 + x/2)^(3/7)*sin(1/3 + 2*x/3), x, 1, 2)1.285098118992018756434479…integral(x^4*sin(1 + x^3/2), x, 1, 2)-1.812892710574357659353896…integral(x^(1/3)*sin(x)*cos(x), x, 1, 3)-0.494468890224824612997499…integral(sin(x^3)^2, x, 1, 2)0.578821039202935718900446…The suite passes (16,934) and so do the MCP tests (329). The new tests differentiate six shapes back, and pin
x^2 sin(x)andsin(x)/xto their old answers.🤖 Generated with Claude Code
https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura