How can I evaluate this integralHow 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
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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
$begingroup$
$$int_-infty^inftyfracsin^2 2x1+x^2operatornamedx$$
I don't know how to solve. I calculated using Wolfram and get
$$fracpisinh 2e^2$$
calculus integration
edited 5 hours ago
J.G.
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
$endgroup$
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
edited 5 hours ago
J.G.
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
$endgroup$
$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
@J.G just a minor remark it is better to usemathopdxto get the correct spacing.
$endgroup$
– zwim
5 hours ago
$begingroup$
@zwim I've settled instead onoperatornamedx.
$endgroup$
– J.G.
5 hours ago
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
edited 5 hours ago
J.G.
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
$endgroup$
$$int_-infty^inftyfracsin^2 2x1+x^2operatornamedx$$
I don't know how to solve. I calculated using Wolfram and get
$$fracpisinh 2e^2$$
calculus integration
calculus integration
edited 5 hours ago
J.G.
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
edited 5 hours ago
J.G.
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
edited 5 hours ago
J.G.
35.9k23555
edited 5 hours ago
J.G.
35.9k23555
edited 5 hours ago
J.G.
35.9k23555
35.9k23555
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
asked 5 hours ago
Pankaj SolankiPankaj Solanki
738
738
$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
@J.G just a minor remark it is better to usemathopdxto get the correct spacing.
$endgroup$
– zwim
5 hours ago
$begingroup$
@zwim I've settled instead onoperatornamedx.
$endgroup$
– J.G.
5 hours ago
add a comment |
$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
@J.G just a minor remark it is better to usemathopdxto get the correct spacing.
$endgroup$
– zwim
5 hours ago
$begingroup$
@zwim I've settled instead onoperatornamedx.
$endgroup$
– J.G.
5 hours ago
$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
@J.G just a minor remark it is better to use
mathopdx to get the correct spacing.$endgroup$
– zwim
5 hours ago
$begingroup$
@J.G just a minor remark it is better to use
mathopdx to get the correct spacing.$endgroup$
– zwim
5 hours ago
$begingroup$
@zwim I've settled instead on
operatornamedx.$endgroup$
– J.G.
5 hours ago
$begingroup$
@zwim I've settled instead on
operatornamedx.$endgroup$
– J.G.
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$
answered 5 hours ago
J.G.J.G.
35.9k23555
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$
answered 5 hours ago
J.G.J.G.
35.9k23555
$endgroup$
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 $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$
answered 5 hours ago
J.G.J.G.
35.9k23555
$endgroup$
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 $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$
answered 5 hours ago
J.G.J.G.
35.9k23555
$endgroup$
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 $$fracpi2e^2left(e^2-e^-2right)=fracpisinh 2e^2.$$
answered 5 hours ago
J.G.J.G.
35.9k23555
edited 5 hours ago
answered 5 hours ago
J.G.J.G.
35.9k23555
answered 5 hours ago
J.G.J.G.
35.9k23555
answered 5 hours ago
J.G.J.G.
35.9k23555
35.9k23555
add a comment |
add a comment |
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$begingroup$
You could do this with contour integration, if you know complex analysis, perhaps. But I'm sure there are other ways.
$endgroup$
– The Count
5 hours ago
$begingroup$
I've changed it to MathJax for you; you can learn how it works from here.
$endgroup$
– J.G.
5 hours ago
$begingroup$
@J.G just a minor remark it is better to use
mathopdxto get the correct spacing.$endgroup$
– zwim
5 hours ago
$begingroup$
@zwim I've settled instead on
operatornamedx.$endgroup$
– J.G.
5 hours ago