Evaluate $int_0^infty cos(bx)(x-ln(e^x-1))dx $

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I have been given the integral
$$int_0^infty cos(bx)(x-ln(e^x-1))dx$$
from a friend. I found an answer in terms of the digamma function, but he told me that the answer is obtainable without imaginary numbers. I am completely dumbfounded on how he got the answer.




I have absolutely no idea where to start without using complex numbers. When $b=0$, it's pretty easy to show that it evaluates to $zeta(2)$ or $fracpi^26$. But I can't figure out a general form without using digamma.




integration




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  • I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
    – Frpzzd
    Aug 7 at 18:50










  • @Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
    – Tom Himler
    Aug 7 at 19:21










  • Very closely related.
    – Simply Beautiful Art
    Aug 7 at 19:22














up vote
5
down vote

favorite
1













I have been given the integral
$$int_0^infty cos(bx)(x-ln(e^x-1))dx$$
from a friend. I found an answer in terms of the digamma function, but he told me that the answer is obtainable without imaginary numbers. I am completely dumbfounded on how he got the answer.




I have absolutely no idea where to start without using complex numbers. When $b=0$, it's pretty easy to show that it evaluates to $zeta(2)$ or $fracpi^26$. But I can't figure out a general form without using digamma.




integration




share|cite|improve this question





















  • I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
    – Frpzzd
    Aug 7 at 18:50










  • @Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
    – Tom Himler
    Aug 7 at 19:21










  • Very closely related.
    – Simply Beautiful Art
    Aug 7 at 19:22












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1






I have been given the integral
$$int_0^infty cos(bx)(x-ln(e^x-1))dx$$
from a friend. I found an answer in terms of the digamma function, but he told me that the answer is obtainable without imaginary numbers. I am completely dumbfounded on how he got the answer.




I have absolutely no idea where to start without using complex numbers. When $b=0$, it's pretty easy to show that it evaluates to $zeta(2)$ or $fracpi^26$. But I can't figure out a general form without using digamma.




integration




share|cite|improve this question














I have been given the integral
$$int_0^infty cos(bx)(x-ln(e^x-1))dx$$
from a friend. I found an answer in terms of the digamma function, but he told me that the answer is obtainable without imaginary numbers. I am completely dumbfounded on how he got the answer.




I have absolutely no idea where to start without using complex numbers. When $b=0$, it's pretty easy to show that it evaluates to $zeta(2)$ or $fracpi^26$. But I can't figure out a general form without using digamma.





integration





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edited Aug 7 at 18:42









Math Lover

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asked Aug 7 at 18:34









Tom Himler

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654213











  • I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
    – Frpzzd
    Aug 7 at 18:50










  • @Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
    – Tom Himler
    Aug 7 at 19:21










  • Very closely related.
    – Simply Beautiful Art
    Aug 7 at 19:22
















  • I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
    – Frpzzd
    Aug 7 at 18:50










  • @Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
    – Tom Himler
    Aug 7 at 19:21










  • Very closely related.
    – Simply Beautiful Art
    Aug 7 at 19:22















I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
– Frpzzd
Aug 7 at 18:50




I am confused... you want to derive a representation without the Digamma function, or derive the Digamma representation without using complex methods?
– Frpzzd
Aug 7 at 18:50












@Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
– Tom Himler
Aug 7 at 19:21




@Frpzzd Sorry if it wasn't clear. He specifically said that "it can be evaluated without the Digamma function or complex numbers."
– Tom Himler
Aug 7 at 19:21












Very closely related.
– Simply Beautiful Art
Aug 7 at 19:22




Very closely related.
– Simply Beautiful Art
Aug 7 at 19:22










1 Answer
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beginalign
int_0^infty cos bx(x-ln(e^x-1))dx
&= int_0^infty cos bxlndfrac11-e^-x dx\
&= int_0^infty cos bxsum_n=1^inftydfrace^-nxn dx\
&= sum_n=1^inftydfrac1nint_0^infty cos bx e^-nx dx\
&= sum_n=1^inftydfrac1ndfracnb^2+n^2\
&= sum_n=1^inftydfrac1b^2+n^2\
&= fracpicothpi b2b-dfrac12b^2
endalign
where $sum_n = 1^infty frac1n^2 + a^2 = fracpicoth(pi a)2a - frac12a^2$. Note that
$$int_0^infty cos bx e^-nx dx=cal L(cos bx)Big|_s=n=dfracnb^2+n^2$$






share|cite|improve this answer





















  • How might one prove the value of that series without using complex methods?
    – Frpzzd
    Aug 7 at 19:04










  • @Frpzzd Fourier series
    – Cyclohexanol.
    Aug 7 at 19:15










  • This is exactly how he did it, thanks!
    – Tom Himler
    Aug 7 at 19:22










  • You are welcome and good luck.
    – Nosrati
    Aug 7 at 19:23











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
7
down vote



accepted










beginalign
int_0^infty cos bx(x-ln(e^x-1))dx
&= int_0^infty cos bxlndfrac11-e^-x dx\
&= int_0^infty cos bxsum_n=1^inftydfrace^-nxn dx\
&= sum_n=1^inftydfrac1nint_0^infty cos bx e^-nx dx\
&= sum_n=1^inftydfrac1ndfracnb^2+n^2\
&= sum_n=1^inftydfrac1b^2+n^2\
&= fracpicothpi b2b-dfrac12b^2
endalign
where $sum_n = 1^infty frac1n^2 + a^2 = fracpicoth(pi a)2a - frac12a^2$. Note that
$$int_0^infty cos bx e^-nx dx=cal L(cos bx)Big|_s=n=dfracnb^2+n^2$$






share|cite|improve this answer





















  • How might one prove the value of that series without using complex methods?
    – Frpzzd
    Aug 7 at 19:04










  • @Frpzzd Fourier series
    – Cyclohexanol.
    Aug 7 at 19:15










  • This is exactly how he did it, thanks!
    – Tom Himler
    Aug 7 at 19:22










  • You are welcome and good luck.
    – Nosrati
    Aug 7 at 19:23















up vote
7
down vote



accepted










beginalign
int_0^infty cos bx(x-ln(e^x-1))dx
&= int_0^infty cos bxlndfrac11-e^-x dx\
&= int_0^infty cos bxsum_n=1^inftydfrace^-nxn dx\
&= sum_n=1^inftydfrac1nint_0^infty cos bx e^-nx dx\
&= sum_n=1^inftydfrac1ndfracnb^2+n^2\
&= sum_n=1^inftydfrac1b^2+n^2\
&= fracpicothpi b2b-dfrac12b^2
endalign
where $sum_n = 1^infty frac1n^2 + a^2 = fracpicoth(pi a)2a - frac12a^2$. Note that
$$int_0^infty cos bx e^-nx dx=cal L(cos bx)Big|_s=n=dfracnb^2+n^2$$






share|cite|improve this answer





















  • How might one prove the value of that series without using complex methods?
    – Frpzzd
    Aug 7 at 19:04










  • @Frpzzd Fourier series
    – Cyclohexanol.
    Aug 7 at 19:15










  • This is exactly how he did it, thanks!
    – Tom Himler
    Aug 7 at 19:22










  • You are welcome and good luck.
    – Nosrati
    Aug 7 at 19:23













up vote
7
down vote



accepted







up vote
7
down vote



accepted






beginalign
int_0^infty cos bx(x-ln(e^x-1))dx
&= int_0^infty cos bxlndfrac11-e^-x dx\
&= int_0^infty cos bxsum_n=1^inftydfrace^-nxn dx\
&= sum_n=1^inftydfrac1nint_0^infty cos bx e^-nx dx\
&= sum_n=1^inftydfrac1ndfracnb^2+n^2\
&= sum_n=1^inftydfrac1b^2+n^2\
&= fracpicothpi b2b-dfrac12b^2
endalign
where $sum_n = 1^infty frac1n^2 + a^2 = fracpicoth(pi a)2a - frac12a^2$. Note that
$$int_0^infty cos bx e^-nx dx=cal L(cos bx)Big|_s=n=dfracnb^2+n^2$$






share|cite|improve this answer













beginalign
int_0^infty cos bx(x-ln(e^x-1))dx
&= int_0^infty cos bxlndfrac11-e^-x dx\
&= int_0^infty cos bxsum_n=1^inftydfrace^-nxn dx\
&= sum_n=1^inftydfrac1nint_0^infty cos bx e^-nx dx\
&= sum_n=1^inftydfrac1ndfracnb^2+n^2\
&= sum_n=1^inftydfrac1b^2+n^2\
&= fracpicothpi b2b-dfrac12b^2
endalign
where $sum_n = 1^infty frac1n^2 + a^2 = fracpicoth(pi a)2a - frac12a^2$. Note that
$$int_0^infty cos bx e^-nx dx=cal L(cos bx)Big|_s=n=dfracnb^2+n^2$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 7 at 18:56









Nosrati

20.2k41644




20.2k41644











  • How might one prove the value of that series without using complex methods?
    – Frpzzd
    Aug 7 at 19:04










  • @Frpzzd Fourier series
    – Cyclohexanol.
    Aug 7 at 19:15










  • This is exactly how he did it, thanks!
    – Tom Himler
    Aug 7 at 19:22










  • You are welcome and good luck.
    – Nosrati
    Aug 7 at 19:23

















  • How might one prove the value of that series without using complex methods?
    – Frpzzd
    Aug 7 at 19:04










  • @Frpzzd Fourier series
    – Cyclohexanol.
    Aug 7 at 19:15










  • This is exactly how he did it, thanks!
    – Tom Himler
    Aug 7 at 19:22










  • You are welcome and good luck.
    – Nosrati
    Aug 7 at 19:23
















How might one prove the value of that series without using complex methods?
– Frpzzd
Aug 7 at 19:04




How might one prove the value of that series without using complex methods?
– Frpzzd
Aug 7 at 19:04












@Frpzzd Fourier series
– Cyclohexanol.
Aug 7 at 19:15




@Frpzzd Fourier series
– Cyclohexanol.
Aug 7 at 19:15












This is exactly how he did it, thanks!
– Tom Himler
Aug 7 at 19:22




This is exactly how he did it, thanks!
– Tom Himler
Aug 7 at 19:22












You are welcome and good luck.
– Nosrati
Aug 7 at 19:23





You are welcome and good luck.
– Nosrati
Aug 7 at 19:23













 

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