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Evaluation of definite integrals with exponential integrand in Fourier sine transform [on hold]

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0

$begingroup$

I was going through a question on finding the Fourier sine transform of:
$$frace^ax+e^-axe^pi x-e^-pi x$$

So I got stuck with this integral:

$$int_0^infty frace^(a+ip)x-e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 tan fraca+ip2$$

The second one which i guess must be quite similar to the former that I encountered in another similar question is:

$$int_0^infty frace^(a+ip)x+e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 sec fraca+ip2$$

I am not able to understand how to proceed with these two. Any help would be appreciated.

definite-integrals exponential-function improper-integrals problem-solving fourier-transform

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edited Mar 29 at 8:59

Shatabdi Sinha

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

$endgroup$

put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, Thomas Shelby, Eevee Trainer, Cesareo Mar 29 at 5:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Thomas Shelby, Eevee Trainer, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding the Fourier transform, see here.
  $endgroup$
  – Maxim
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

I was going through a question on finding the Fourier sine transform of:
$$frace^ax+e^-axe^pi x-e^-pi x$$

So I got stuck with this integral:

$$int_0^infty frace^(a+ip)x-e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 tan fraca+ip2$$

The second one which i guess must be quite similar to the former that I encountered in another similar question is:

$$int_0^infty frace^(a+ip)x+e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 sec fraca+ip2$$

I am not able to understand how to proceed with these two. Any help would be appreciated.

definite-integrals exponential-function improper-integrals problem-solving fourier-transform

share|cite|improve this question

edited Mar 29 at 8:59

Shatabdi Sinha

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

$endgroup$

put on hold as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, Thomas Shelby, Eevee Trainer, Cesareo Mar 29 at 5:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Thomas Shelby, Eevee Trainer, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Regarding the Fourier transform, see here.
  $endgroup$
  – Maxim
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

I was going through a question on finding the Fourier sine transform of:
$$frace^ax+e^-axe^pi x-e^-pi x$$

So I got stuck with this integral:

$$int_0^infty frace^(a+ip)x-e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 tan fraca+ip2$$

The second one which i guess must be quite similar to the former that I encountered in another similar question is:

$$int_0^infty frace^(a+ip)x+e^-(a+ip)xdxe^pi x-e^-pi x$$
$$= frac12 sec fraca+ip2$$

I am not able to understand how to proceed with these two. Any help would be appreciated.

definite-integrals exponential-function improper-integrals problem-solving fourier-transform

share|cite|improve this question

edited Mar 29 at 8:59

Shatabdi Sinha

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

$endgroup$

share|cite|improve this question

edited Mar 29 at 8:59

Shatabdi Sinha

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

share|cite|improve this question

edited Mar 29 at 8:59

Shatabdi Sinha

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

asked Mar 28 at 15:46

Shatabdi SinhaShatabdi Sinha

19113

1 Answer
1

active

oldest

votes

2

$begingroup$

Note that we can write for $|a|<pi$

$$beginalign
int_0^infty frace^(a+ip)x-e^-(a+ip)xleft(e^pi x-e^-pi xright),dx&=int_0^infty frace^-pi xleft(e^(a+ip)x-e^-(a+ip)xright)1-e^-2pi x,dx\\
&=sum_n=0^infty int_0^infty e^-(2n+1)pi xleft(e^(a+ip)x-e^-(a+ip)xright)\\
&=sum_n=0^inftyleft(frac1(2n+1)pi -(a+ip)-frac1(2n+1)pi +(a+ip)right)\\
&=-frac12sum_n=-infty^infty frac1fraca+ip2+fracpi2+npitag1\\
&=frac12tanleft(fraca+ip2right)tag2
endalign$$

where in going from $(1)$ to $(2)$ we noted that right-hand side of $(1)$ was the partial fraction representation of $-cotleft(fraca+ip2+fracpi2right)=tanleft(fraca+ip2right)$ (See THIS ANSWER and the Appendix of THIS ONE).

share|cite|improve this answer

edited Mar 28 at 17:57

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Partial fraction expansion representation of the tangent function. Could you please elaborate on this. That would be really helpful.
  $endgroup$
  – Shatabdi Sinha
  Mar 28 at 16:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. See THIS and the Appendix of THIS answer.
  $endgroup$
  – Mark Viola
  Mar 28 at 17:26
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that we can write for $|a|<pi$

$$beginalign
int_0^infty frace^(a+ip)x-e^-(a+ip)xleft(e^pi x-e^-pi xright),dx&=int_0^infty frace^-pi xleft(e^(a+ip)x-e^-(a+ip)xright)1-e^-2pi x,dx\\
&=sum_n=0^infty int_0^infty e^-(2n+1)pi xleft(e^(a+ip)x-e^-(a+ip)xright)\\
&=sum_n=0^inftyleft(frac1(2n+1)pi -(a+ip)-frac1(2n+1)pi +(a+ip)right)\\
&=-frac12sum_n=-infty^infty frac1fraca+ip2+fracpi2+npitag1\\
&=frac12tanleft(fraca+ip2right)tag2
endalign$$

where in going from $(1)$ to $(2)$ we noted that right-hand side of $(1)$ was the partial fraction representation of $-cotleft(fraca+ip2+fracpi2right)=tanleft(fraca+ip2right)$ (See THIS ANSWER and the Appendix of THIS ONE).

share|cite|improve this answer

edited Mar 28 at 17:57

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Partial fraction expansion representation of the tangent function. Could you please elaborate on this. That would be really helpful.
  $endgroup$
  – Shatabdi Sinha
  Mar 28 at 16:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. See THIS and the Appendix of THIS answer.
  $endgroup$
  – Mark Viola
  Mar 28 at 17:26
  

  
  
  
  

add a comment | 

2

$begingroup$

Note that we can write for $|a|<pi$

$$beginalign
int_0^infty frace^(a+ip)x-e^-(a+ip)xleft(e^pi x-e^-pi xright),dx&=int_0^infty frace^-pi xleft(e^(a+ip)x-e^-(a+ip)xright)1-e^-2pi x,dx\\
&=sum_n=0^infty int_0^infty e^-(2n+1)pi xleft(e^(a+ip)x-e^-(a+ip)xright)\\
&=sum_n=0^inftyleft(frac1(2n+1)pi -(a+ip)-frac1(2n+1)pi +(a+ip)right)\\
&=-frac12sum_n=-infty^infty frac1fraca+ip2+fracpi2+npitag1\\
&=frac12tanleft(fraca+ip2right)tag2
endalign$$

where in going from $(1)$ to $(2)$ we noted that right-hand side of $(1)$ was the partial fraction representation of $-cotleft(fraca+ip2+fracpi2right)=tanleft(fraca+ip2right)$ (See THIS ANSWER and the Appendix of THIS ONE).

share|cite|improve this answer

edited Mar 28 at 17:57

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Partial fraction expansion representation of the tangent function. Could you please elaborate on this. That would be really helpful.
  $endgroup$
  – Shatabdi Sinha
  Mar 28 at 16:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Sure. See THIS and the Appendix of THIS answer.
  $endgroup$
  – Mark Viola
  Mar 28 at 17:26
  

  
  
  
  

add a comment | 

$begingroup$

Note that we can write for $|a|<pi$

$$beginalign
int_0^infty frace^(a+ip)x-e^-(a+ip)xleft(e^pi x-e^-pi xright),dx&=int_0^infty frace^-pi xleft(e^(a+ip)x-e^-(a+ip)xright)1-e^-2pi x,dx\\
&=sum_n=0^infty int_0^infty e^-(2n+1)pi xleft(e^(a+ip)x-e^-(a+ip)xright)\\
&=sum_n=0^inftyleft(frac1(2n+1)pi -(a+ip)-frac1(2n+1)pi +(a+ip)right)\\
&=-frac12sum_n=-infty^infty frac1fraca+ip2+fracpi2+npitag1\\
&=frac12tanleft(fraca+ip2right)tag2
endalign$$

where in going from $(1)$ to $(2)$ we noted that right-hand side of $(1)$ was the partial fraction representation of $-cotleft(fraca+ip2+fracpi2right)=tanleft(fraca+ip2right)$ (See THIS ANSWER and the Appendix of THIS ONE).

share|cite|improve this answer

edited Mar 28 at 17:57

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

edited Mar 28 at 17:57

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176

answered Mar 28 at 16:26

Mark ViolaMark Viola

134k1278176