transform integral of a series












0














I am given the following:




$1. int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt$ where $sin mathbb R$




How do I derive the following form of the integral above ?



$$sum_{n=0}^{infty} frac{1}{(n^2pi)^s}int_0^{infty}exp(-t)t^{s-1}dt$$



I tried to rearrange $(1)$ and use partial integration but that didn't work out.



Would appreciate any help/hints






integration sequences-and-series







share|cite|improve this question




















  • 2




    You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
    – Lord Shark the Unknown
    Oct 23 at 15:56






  • 2




    Please fix the typo(s).
    – Yves Daoust
    Oct 23 at 16:04
















0














I am given the following:




$1. int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt$ where $sin mathbb R$




How do I derive the following form of the integral above ?



$$sum_{n=0}^{infty} frac{1}{(n^2pi)^s}int_0^{infty}exp(-t)t^{s-1}dt$$



I tried to rearrange $(1)$ and use partial integration but that didn't work out.



Would appreciate any help/hints






integration sequences-and-series







share|cite|improve this question




















  • 2




    You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
    – Lord Shark the Unknown
    Oct 23 at 15:56






  • 2




    Please fix the typo(s).
    – Yves Daoust
    Oct 23 at 16:04














0












0








0







I am given the following:




$1. int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt$ where $sin mathbb R$




How do I derive the following form of the integral above ?



$$sum_{n=0}^{infty} frac{1}{(n^2pi)^s}int_0^{infty}exp(-t)t^{s-1}dt$$



I tried to rearrange $(1)$ and use partial integration but that didn't work out.



Would appreciate any help/hints






integration sequences-and-series







share|cite|improve this question















I am given the following:




$1. int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt$ where $sin mathbb R$




How do I derive the following form of the integral above ?



$$sum_{n=0}^{infty} frac{1}{(n^2pi)^s}int_0^{infty}exp(-t)t^{s-1}dt$$



I tried to rearrange $(1)$ and use partial integration but that didn't work out.



Would appreciate any help/hints






integration sequences-and-series




integration sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 16:13









Mostafa Ayaz

13.7k3836




13.7k3836










asked Oct 23 at 15:54









XPenguen

963617




963617








  • 2




    You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
    – Lord Shark the Unknown
    Oct 23 at 15:56






  • 2




    Please fix the typo(s).
    – Yves Daoust
    Oct 23 at 16:04














  • 2




    You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
    – Lord Shark the Unknown
    Oct 23 at 15:56






  • 2




    Please fix the typo(s).
    – Yves Daoust
    Oct 23 at 16:04








2




2




You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
– Lord Shark the Unknown
Oct 23 at 15:56




You surely want $exp(-n^2pi t)$ in the first integral. Now make the substitution $t'=-n^2pi t$.
– Lord Shark the Unknown
Oct 23 at 15:56




2




2




Please fix the typo(s).
– Yves Daoust
Oct 23 at 16:04




Please fix the typo(s).
– Yves Daoust
Oct 23 at 16:04










1 Answer
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Thanks to @LordSharkTheUnknown by substituting $t'={ n^2pi }t$ we obtain$$int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt{=int_0^{infty}sum_{n=0}^{infty}exp(-t')left({t'over n^2pi}right)^{s-1}d({t'over n^2pi})\=int_0^{infty}sum_{n=0}^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}int_0^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}left({1over n^2pi }right)^sint_0^{infty}exp(-t)cdot t^{s-1}dt}$$






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    Thanks to @LordSharkTheUnknown by substituting $t'={ n^2pi }t$ we obtain$$int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt{=int_0^{infty}sum_{n=0}^{infty}exp(-t')left({t'over n^2pi}right)^{s-1}d({t'over n^2pi})\=int_0^{infty}sum_{n=0}^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}int_0^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}left({1over n^2pi }right)^sint_0^{infty}exp(-t)cdot t^{s-1}dt}$$






    share|cite|improve this answer


























      1














      Thanks to @LordSharkTheUnknown by substituting $t'={ n^2pi }t$ we obtain$$int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt{=int_0^{infty}sum_{n=0}^{infty}exp(-t')left({t'over n^2pi}right)^{s-1}d({t'over n^2pi})\=int_0^{infty}sum_{n=0}^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}int_0^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}left({1over n^2pi }right)^sint_0^{infty}exp(-t)cdot t^{s-1}dt}$$






      share|cite|improve this answer
























        1












        1








        1






        Thanks to @LordSharkTheUnknown by substituting $t'={ n^2pi }t$ we obtain$$int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt{=int_0^{infty}sum_{n=0}^{infty}exp(-t')left({t'over n^2pi}right)^{s-1}d({t'over n^2pi})\=int_0^{infty}sum_{n=0}^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}int_0^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}left({1over n^2pi }right)^sint_0^{infty}exp(-t)cdot t^{s-1}dt}$$






        share|cite|improve this answer












        Thanks to @LordSharkTheUnknown by substituting $t'={ n^2pi }t$ we obtain$$int_0^{infty}sum_{n=0}^{infty}exp(-n^2pi t)t^{s-1}dt{=int_0^{infty}sum_{n=0}^{infty}exp(-t')left({t'over n^2pi}right)^{s-1}d({t'over n^2pi})\=int_0^{infty}sum_{n=0}^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}int_0^{infty}left({1over n^2pi }right)^sexp(-t'){(t')}^{s-1}dt'\=sum_{n=0}^{infty}left({1over n^2pi }right)^sint_0^{infty}exp(-t)cdot t^{s-1}dt}$$







        share|cite|improve this answer












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        answered Nov 30 at 16:21









        Mostafa Ayaz

        13.7k3836




        13.7k3836






























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