1
$\begingroup$

Do you guys have any ideas on solving the following definite integral: $$\int_{0}^{R} x^{\lambda} (1-x)^{\mu-1} K_{1}(a\sqrt{x}) \, dx$$ where the parameters $\lambda>0$, $0 < \mu <1$, $a>0$, and $K_{1}(\cdot)$ is the modified Bessel function of second kind with order 1.

I tried to conduct the integration by parts but eventually the (1-x) term will just not vanish. I also found the following definite integral from the book `Table of Integrals, Series and Products´. However, it is not straightforward for me to derive the antiderivative from the definite integral enter image description here

$\endgroup$
  • $\begingroup$ WolframMathematica doesnt give any result. This is the code I had used Integrate[x^\[Lambda] (1 - x)^(\[Mu] - 1) BesselI[1, a Sqrt[x]], x, Assumptions -> \[Lambda] > 0 && 0 < \[Mu] && a > 0] $\endgroup$ – Masacroso Apr 7 '18 at 1:20
  • $\begingroup$ @Masacroso, I have also tried Mathematica when I tried to solve the problem and did not get results, $\endgroup$ – Vic Apr 7 '18 at 1:26
  • $\begingroup$ It might not have any kind of closed form. Even the definite integral you provided is not that useful, becuase it uses generalized functions. I would just use numerical methods. I will try to get something later though (like in your other question), maybe there's a way $\endgroup$ – Yuriy S Apr 7 '18 at 10:30
  • $\begingroup$ @Yuriy S, To make it easier, I have now modified the question a little bit from a indefinite integral to definite integral. Do you think it is possible to solve it with generalized functions? Any solution using generalized function would also be great. Today I tried to solve it by expressing the Bessel function with series expansion, I got an answer but it works only with small R as the series expansion is only valid close to 0. $\endgroup$ – Vic Apr 7 '18 at 19:58
  • $\begingroup$ Hi @YuriyS, I just found the following way to solve the new definite integral $\int_{0}^{R} x^{\lambda} (1-x)^{\mu-1} K_{1}(a\sqrt{x}) \, dx$, we can use the change of integration interval, i.e., $\int_{0}^{R} f(x)\,dx = R \int_{0}^{1} f(R x)\,dx$, where the solution to the integral $\int_{0}^{1} f(\cdot )\,dx$ has been given in terms of Meijer G function. Do you think this is the right way to obtain the solution? $\endgroup$ – Vic Apr 8 '18 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.