calculus - Integral $\int (t^2 - 1)^{a} \cdot t^{b} \cdot \log(t)\,dt$ - Mathematics Stack Exchange
the integral $\int (t^2 - 1)^{a} \cdot t^{b} \cdot \log(t)\,dt$ has following solution in terms of hypergeometric functions according wolfram:
the solution has been check extensively. have tried use integration parts, did not find references on integral leading hypergeometric function resources.
does know how final solution in terms of 2 hypergeometric functions?
i got answer in terms of hypergeometric functions correct numerically , can transformed answer wolfram gives.
- the answer (i omit constant):
$$\int t^b (t^2-1)^a \ln t~dt= \\ = \frac{t^{2a+b+1}}{2a+b+1} \bigg( (\ln t) {_2 f_1} \left(-a,-a-\frac{b+1}{2};-a-\frac{b-1}{2}; \frac{1}{t^2} \right) - \\ - \frac{1}{2a+b+1} {_3 f_2} \left(-a,-a-\frac{b+1}{2},-a-\frac{b+1}{2};-a-\frac{b-1}{2},-a-\frac{b-1}{2}; \frac{1}{t^2} \right) \bigg)$$
- the formulas used (per wikipedia hypergeometric function , generalized hypergeometric function):
$$\mathrm {b} (b,c-b)\,_{2}f_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx$$
$$_{a+1}f_{b+1}\left[{\begin{array}{c}a_{1},\ldots ,a_{a},c\\b_{1},\ldots ,b_{b},d\end{array}};z\right]={\frac {\gamma (d)}{\gamma (c)\gamma (d-c)}}\int _{0}^{1}t^{c-1}(1-t)_{}^{d-c-1}\ {}_{a}f_{b}\left[{\begin{array}{c}a_{1},\ldots ,a_{a}\\b_{1},\ldots ,b_{b}\end{array}};tz\right]dt$$
the solution.
- first, find integral (i have chosen limits can trace substitutions, lower, constant limit can arbitrary):
$$\int^{t_0}_1 t^b (t^2-1)^a~dt$$
we can changing variable following way:
$$\frac{1}{t} = x \quad \to \quad x^2=y \quad \to \quad y=\frac{z}{t_0^2}$$
if perform every substitution correctly, find 2 integrals in form given wikipedia:
$$\int^{t_0}_1 t^b (t^2-1)^a~dt= \\ = - \frac{t_0^{2a+b+1}}{2} \int_0^1 \left(1-\frac{1}{t_0^2} z \right)^a z^{-a-(b+3)/2} dz+\frac{1}{2} \int_0^1 \left(1- y \right)^a y^{-a-(b+3)/2} dy$$
the last expression constant, first 1 gives hypergeometric function (you need brush on properties of beta function):
$$\int^{t_0}_1 t^b (t^2-1)^a~dt= \\ = \frac{t_0^{2a+b+1}}{2a+b+1} {_2 f_1} \left(-a,-a-\frac{b+1}{2};-a-\frac{b-1}{2}; \frac{1}{t_0^2} \right)+\text{const}$$
- now use integration parts formula:
$$\int t^b (t^2-1)^a \ln t~dt= \ln t~\int t^b (t^2-1)^a ~dt-\int \frac{1}{t} \int t_1^b (t_1^2-1)^a ~dt_1~dt$$
- we need find integral:
$$\int t^{2a+b} {_2 f_1} \left(-a,-a-\frac{b+1}{2};-a-\frac{b-1}{2}; \frac{1}{t^2} \right) dt$$
we use same substitutions above transform integral form required second formula wikipedia , obtain answer.
here's example of numerical confirmation expression correct:
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