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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:

enter image description here

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?

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  • $\begingroup$ The first thing that stands out is that$$\frac {\partial}{\partial b}t^b=t^b\log t$$So we can write your integral as$$\mathcal{I}=\frac {\partial}{\partial b}\int dt\,(t^2-1)^a t^b$$ $\endgroup$ – Crescendo Jan 20 '18 at 19:36
  • $\begingroup$ @Vic Not quite $$\frac{\partial}{\partial t}t^b = bt^{b-1}$$ $\endgroup$ – Dylan Jan 20 '18 at 20:11
  • $\begingroup$ @Dylan, I have done some search on the Feynman's trick, but it is not that obvious to me, could you elaborate on using Feynman's trick in this problem. $\endgroup$ – Vic Jan 20 '18 at 20:14
  • $\begingroup$ Define $F(b)$ as your integral. Then you have $F'(b)$ as the second integral in the above comment. You'll also need to find $F(0)$ by integrating $$ \int (t^2-1)^a \ln t\ dt $$. Then just integrate back $F'(b)$ and find the constant $\endgroup$ – Dylan Jan 20 '18 at 20:17
  • $\begingroup$ I don't have any hints for you. I'm just elaborating on the first comment. $\endgroup$ – Dylan Jan 20 '18 at 20:18
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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):

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

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

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

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

  1. 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.


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

enter image description here

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  • $\begingroup$ There was a typo in the answer, of course it's ${_3 F_2}$ $\endgroup$ – Yuriy S Jan 21 '18 at 2:22
  • $\begingroup$ that's really brilliant, we usually solve definite integral by using its antiderivative, and here it goes the opposite way, thumbs-up :) $\endgroup$ – Vic Jan 21 '18 at 9:22
  • $\begingroup$ may I know which property of Beta function have you used to obtain the final answer, as Beta function is linked with Gamma function and it seems that there is no direct relationship between the Beta function and the expression (2a+b+1). Thanks a lot. $\endgroup$ – Vic Apr 4 '18 at 11:15
  • $\begingroup$ @Vic, I think it's the property $B(1,x)=\frac{1}{x}$, thought I'm not sure, I don't remember the details of this calculation $\endgroup$ – Yuriy S Apr 4 '18 at 11:21
  • $\begingroup$ yep, that's it and everything is crystal clear now, thanks a lot. $\endgroup$ – Vic Apr 5 '18 at 15:27

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