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:

enter image description here

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.

  1. the answer (i omit constant):

$$\int t^b (t^2-1)^a \ln t~dt= \\ = \frac{t^{2a+b+1}}{2a+b+1} \bigg( (\ln t) {_2 f_1} \left(-a,-a-\frac{b+1}{2};-a-\frac{b-1}{2}; \frac{1}{t^2} \right) - \\ - \frac{1}{2a+b+1} {_3 f_2} \left(-a,-a-\frac{b+1}{2},-a-\frac{b+1}{2};-a-\frac{b-1}{2},-a-\frac{b-1}{2}; \frac{1}{t^2} \right) \bigg)$$

  1. the formulas used (per wikipedia hypergeometric function , generalized hypergeometric function):

$$\mathrm {b} (b,c-b)\,_{2}f_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx$$

$$_{a+1}f_{b+1}\left[{\begin{array}{c}a_{1},\ldots ,a_{a},c\\b_{1},\ldots ,b_{b},d\end{array}};z\right]={\frac {\gamma (d)}{\gamma (c)\gamma (d-c)}}\int _{0}^{1}t^{c-1}(1-t)_{}^{d-c-1}\ {}_{a}f_{b}\left[{\begin{array}{c}a_{1},\ldots ,a_{a}\\b_{1},\ldots ,b_{b}\end{array}};tz\right]dt$$

  1. 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:

enter image description here


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