I have been trying to solve for f(x) by doing the Inverse Mellin Transform of F(s) by summing up residues, but I don't think I'm on the right track.

$$f(x)=\frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}F(s)x^{-s}ds $$


I know that the residues are at $\Gamma(-n)$ so if I solve for s I can find where the poles are. I know that there are poles at s=-4n where n=0,1,2,3... for example. If I can sum up each residue I can solve for f(x).I don't know where the line integral for c should be either. I have tried looking at integral tables and have looked at Mellin-barnes integrals a little bit but I need help.

One thing that confuses me is that there are poles at $s=4n+1$ so I get negative powers when plugging $s$ into $x^{-s}$. Thanks again for the help!

  • $\begingroup$ my guess is that you'll get Bessel functions or the like. does that make sense in relation to the problem at hand? $\endgroup$ – mathstackuser12 Apr 24 '18 at 4:46

This is a very quick sketch - i would need to spend more time on it to sort it out properly, however perhaps this might give you some hints. Firstly using the asymptotic series for the gamma function you can show (i think)

$${{4}^{s/2}}\frac{\Gamma \left( \tfrac{1}{4}s \right)\Gamma \left( \tfrac{3}{4}+\tfrac{1}{4}s \right)\Gamma \left( \tfrac{1}{4}-\tfrac{1}{4}s \right)}{\Gamma \left( \tfrac{3}{4}-\tfrac{1}{4}s \right)}{{x}^{-s}}\sim 2\pi {{e}^{\frac{1}{2}\left( \ln \left( 4 \right)-s\left( 1+\ln \left( {{x}^{2}} \right) \right)+\left( s-1 \right)\ln \left( s \right) \right)}}\left| s \right|\to \infty $$

There could be a typo in there. Eitherway it seems pretty obvious that we have to close the contour to the left and not the right (another way to see this is that if you do close the contour to the right you would be counting residues of the poles at s=4n+1 which leads to a divergent series...i think! again check me.)

So now that we are closing to the left we need to pick up poles, the relevant ones being $$\Gamma \left( \tfrac{1}{4}s \right)\to s=-4n,\,\,res=\frac{{{\left( -1 \right)}^{n}}4}{n!}\,\,\,\,\Gamma \left( \tfrac{3}{4}+\tfrac{1}{4}s \right)\to s=-4n-3,\,\,res=\frac{{{\left( -1 \right)}^{n}}4}{n!}$$.

Plugging in the values of s at the poles and multiplying by the residues we have

$$f\left( x \right)=4\sum\limits_{n=0}^{\infty }{\frac{\Gamma \left( \tfrac{3}{4}-n \right)\Gamma \left( \tfrac{1}{4}+n \right)}{\Gamma \left( \tfrac{3}{4}+n \right)}\frac{{{\left( -1 \right)}^{n}}}{n!}{{\left( \frac{x}{2} \right)}^{4n}}+}\frac{{{x}^{3}}}{2}\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\frac{\Gamma \left( -n-\tfrac{3}{4} \right)}{\Gamma \left( \tfrac{3}{2}+n \right)}{{\left( \frac{x}{2} \right)}^{4n}}}$$

Now if you massage the gamma functions within both series you will end up being able to recognise two functions: the modified Bessel function of the first kind and the modified Struve function;

$$f\left( x \right)=4\pi \sqrt{x}\,\left( {{I}_{-1/4}}\left( {{x}^{2}}/2 \right)-\,{{L}_{1/4}}\left( {{x}^{2}}/2 \right) \right)$$

| cite | improve this answer | |
  • $\begingroup$ Thanks that is very helpful. $\endgroup$ – rms26 Apr 28 '18 at 20:02

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.