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How can I evaluate this integral

How to solve this double integral?Evaluate integralCan this integral be computed in closed-form?How to evaluate this expected value integral?why doesn't wolfram alpha evaluate this integral?How do I can evaluate this integral: $int_1^inftyfracycosh(yx)sinh(ypi)dy$?How to tell if this integral converges?How I can evaluate this integral?How to evaluate this gaussian integralHow do I evaluate this integral

1

2

$begingroup$

$$int_-infty^inftyfracsin^2 2x1+x^2operatornamedx$$

I don't know how to solve. I calculated using Wolfram and get

$$fracpisinh 2e^2$$

calculus integration

share|cite|improve this question

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
  $endgroup$
  – The Count
  May 4 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've changed it to MathJax for you; you can learn how it works from here.
  $endgroup$
  – J.G.
  May 4 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G just a minor remark it is better to use mathopdx to get the correct spacing.
  $endgroup$
  – zwim
  May 4 at 21:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zwim I've settled instead on operatornamedx.
  $endgroup$
  – J.G.
  May 4 at 21:16
  

  
  
  
  

add a comment | 

1

2

$begingroup$

$$int_-infty^inftyfracsin^2 2x1+x^2operatornamedx$$

I don't know how to solve. I calculated using Wolfram and get

$$fracpisinh 2e^2$$

calculus integration

share|cite|improve this question

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
  $endgroup$
  – The Count
  May 4 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've changed it to MathJax for you; you can learn how it works from here.
  $endgroup$
  – J.G.
  May 4 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G just a minor remark it is better to use mathopdx to get the correct spacing.
  $endgroup$
  – zwim
  May 4 at 21:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zwim I've settled instead on operatornamedx.
  $endgroup$
  – J.G.
  May 4 at 21:16
  

  
  
  
  

add a comment | 

$begingroup$

$$int_-infty^inftyfracsin^2 2x1+x^2operatornamedx$$

I don't know how to solve. I calculated using Wolfram and get

$$fracpisinh 2e^2$$

calculus integration

share|cite|improve this question

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

$endgroup$

share|cite|improve this question

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

share|cite|improve this question

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

edited May 4 at 21:16

J.G.

37.5k23656

edited May 4 at 21:16

J.G.

37.5k23656

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

asked May 4 at 21:02

Pankaj SolankiPankaj Solanki

758

1 Answer
1

active

oldest

votes

5

$begingroup$

Your integral is $$frac12int_Bbb Rfrac1-cos 4x1+x^2operatornamedx=fracpi-Reint_Bbb Rfracexp 4ix1+x^2operatornamedx2=fracpi2(1-exp -4),$$using e.g. the Cauchy distribution's characteristic function,$$frac1piint_Bbb Rfracexp itx1+x^2operatornamedx=exp -|t|.$$We can rewrite this result as $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$

share|cite|improve this answer

edited May 4 at 21:15

answered May 4 at 21:10

J.G.J.G.

37.5k23656

$endgroup$

add a comment | 

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5

$begingroup$

Your integral is $$frac12int_Bbb Rfrac1-cos 4x1+x^2operatornamedx=fracpi-Reint_Bbb Rfracexp 4ix1+x^2operatornamedx2=fracpi2(1-exp -4),$$using e.g. the Cauchy distribution's characteristic function,$$frac1piint_Bbb Rfracexp itx1+x^2operatornamedx=exp -|t|.$$We can rewrite this result as $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$

share|cite|improve this answer

edited May 4 at 21:15

answered May 4 at 21:10

J.G.J.G.

37.5k23656

$endgroup$

add a comment | 

5

$begingroup$

Your integral is $$frac12int_Bbb Rfrac1-cos 4x1+x^2operatornamedx=fracpi-Reint_Bbb Rfracexp 4ix1+x^2operatornamedx2=fracpi2(1-exp -4),$$using e.g. the Cauchy distribution's characteristic function,$$frac1piint_Bbb Rfracexp itx1+x^2operatornamedx=exp -|t|.$$We can rewrite this result as $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$

share|cite|improve this answer

edited May 4 at 21:15

answered May 4 at 21:10

J.G.J.G.

37.5k23656

$endgroup$

add a comment | 

$begingroup$

Your integral is $$frac12int_Bbb Rfrac1-cos 4x1+x^2operatornamedx=fracpi-Reint_Bbb Rfracexp 4ix1+x^2operatornamedx2=fracpi2(1-exp -4),$$using e.g. the Cauchy distribution's characteristic function,$$frac1piint_Bbb Rfracexp itx1+x^2operatornamedx=exp -|t|.$$We can rewrite this result as $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$

share|cite|improve this answer

edited May 4 at 21:15

answered May 4 at 21:10

J.G.J.G.

37.5k23656

$endgroup$

share|cite|improve this answer

edited May 4 at 21:15

answered May 4 at 21:10

J.G.J.G.

37.5k23656

answered May 4 at 21:10

J.G.J.G.

37.5k23656

answered May 4 at 21:10

J.G.J.G.

37.5k23656