Integral $\int (t^2 - 1)^{a} \cdot t^{b} \cdot \log(t)\,dt$

Question

The Integral $\int (t^2 - 1)^{a} \cdot t^{b} \cdot \log(t)\,dt$ has the following solution in terms of hypergeometric functions according to Wolfram:

The solution has also been check extensively. I have tried to use integration by parts, but did not find any references on integral leading to the hypergeometric function with my resources.

Does anyone know how to get this final solution in terms of the two hypergeometric functions?

Answer 1

I got another answer in terms of hypergeometric functions which is also correct numerically and can be likely transformed into the answer Wolfram gives.

1. The answer is (I omit the constant):

$$\int t^b (t^2-1)^a \ln t~dt= \\ = \frac{t^{2a+b+1}}{2a+b+1} \bigg( (\ln t) {_2 F_1} \left(-a,-a-\frac{b+1}{2};-a-\frac{b-1}{2}; \frac{1}{t^2} \right) - \\ - \frac{1}{2a+b+1} {_3 F_2} \left(-a,-a-\frac{b+1}{2},-a-\frac{b+1}{2};-a-\frac{b-1}{2},-a-\frac{b-1}{2}; \frac{1}{t^2} \right) \bigg)$$

2. The formulas used (per Wikipedia Hypergeometric function and Generalized hypergeometric function):

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

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

3. The solution.

   • First, we will find the integral (I have chosen the limits so we can easily trace the substitutions, the lower, constant limit can be arbitrary):

$$\int^{t_0}_1 t^b (t^2-1)^a~dt$$

We can do that by changing the variable the following way:

$$\frac{1}{t} = x \quad \to \quad x^2=y \quad \to \quad y=\frac{z}{t_0^2}$$

If you perform every substitution correctly, you will find two integrals in the form given by 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 is just a constant, the first one gives the hypergeometric function (you will also need to brush up on the 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 we use the integration by 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 to find the 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 the same substitutions as above to transform this integral to the form required by the second formula from Wikipedia and obtain the answer.

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Here's an example of numerical confirmation that the expression is correct: