Is there an optimal way to check complicated definite integrals?

[wasia]

Hello!

I am looking for a value of an integral
$$\int^{\infty}_0 {r^{3-\epsilon} \over (r^2+N^2)^2}dr$$

I have tried looking up a book by Gradshteyn and Ryzhik, however, its structure is quite complicated. Should I rewrite the integrand in some other non-obvious way to find it? Would you recommend using some other resource?

The answer is known (it involves Gamma functions), as the integral is a part of a paper about the "ABC theory" (toy QFT) by Kraus and Griffiths. However, I would like to 1) discover the optimal way to check complicated definite integrals in future and 2) check the value of this particular integral.

Thank you.

 

[yyat]

Not sure if this helps, but you can always use the Wolfram online integrator:

http://integrals.wolfram.com/index.jsp?expr=x^(3-e)%2F(x^2+%2B+a^2)^2&random=false

 

[wasia]

Not sure if this helps, but you can always use the Wolfram online integrator:

http://integrals.wolfram.com/index.jsp?expr=x^(3-e)%2F(x^2+%2B+a^2)^2&random=false

I have tried it, this integrator does not do the definite integration. Input of the integrand above gives some "hypergeometric" functions as an output. I still hope that something better exists, but thank you anyway.

Maple or Mathematica would be a possibility, but I usually do not have them at hand.

 

[wasia]

I have failed to find the value in the integral tables, but I post the solution here, in case someone needs it.

$$R= \int_0^\infty {r^{3-\epsilon}dr\over \left[r^2+\Lambda^2\right]^2} = \int_0^\infty {rdr(r^2)^{1-\epsilon /2}dr\over \left[r^2+\Lambda^2\right]^2} = \left[ \substack{r^2+\Lambda^2=\Lambda^2 / y \\ 2rdr = -\Lambda^2 dy / y^2} \right]$$
$$= \int_1^0 \left( -{\Lambda^2dy \over 2y^2 } \right) \left[ \Lambda^2 \left( {1-y\over y} \right) \right]^{1-\epsilon /2} \left( y \over \Lambda^2 \right)^2 ={1 \over 2\Lambda^{\epsilon}} \int_0^1 \left( {1-y \over y} \right)^{1-\epsilon /2}dy.$$

Then we have to know what Euler Beta and Gamma functions are.

$$B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}dt={\Gamma (a)\Gamma (b)\over \Gamma (a+b)}.$$

And finally

$$R = \dfrac{1}{2\Lambda^{\epsilon}} B({\epsilon \over 2}, 2-{\epsilon \over 2}) = {\Gamma(\epsilon /2)\Gamma(2-\epsilon /2) \over 2\Lambda^{\epsilon}\Gamma(2)} = {\Gamma(\epsilon /2)\Gamma(2-\epsilon /2) \over 2\Lambda^{\epsilon}}.$$

 

[bpet]

$$R = {\Gamma(\epsilon /2)\Gamma(2-\epsilon /2) \over 2\Lambda^{\epsilon}}.$$

Using Mathematica I get
$$R = \frac{\pi(2-\epsilon)}{4\Lambda^{\epsilon}\sin(\epsilon\pi/2)}$$
which appears to be equivalent.

 

[Santa1]

Yes by $$\frac{\pi}{\sin(\pi x)}=\Gamma(x)\Gamma(1-x)$$ one gets that result.