How to simplify the result I obtain for the following integral: $int_0^infty frac{cos(a x)}{x^4+b^4},dx$












4












$begingroup$


I have been trying to calculate the following integral using the Residue Theorem but end up with an answer that seems to contain an imaginary part while the integral should be purely real:
$$I_1=int_0^infty frac{cos(a x)}{x^4+b^4},dx$$
In order to calculate this integral, I consider the contour integral
$$I_2=int_C frac{e^{iaz}}{z^4+b^4},dz$$
Where $C$ is the semi-circle of radius $R$ in the upper half plane parameterized counter-clockwise. As $Rtoinfty$, the semi-circular integral vanishes, and a symmetry argument shows that $I_1=I_2/2$.



Now, we can use the Residue Theorem to compute $I_2$, which has poles at $z=be^{ipi/4}$ and $z=be^{3ipi/4}$. After some computations, I obtain
begin{align}I_1&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}+e^{-pi ab /4}e^{-3ipi/4} right)\
&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}-ie^{-pi ab /4}e^{-ipi/4} right) \
&=frac{pi i}{4b^3} e^{-ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
&=frac{pi}{4b^3} e^{ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
end{align}

Now, it seems that unless $a=0$ ( and $b=0$, but at that point the integral is undefined), there will always be a non-zero imaginary part, which shouldn't occur because, as I said earlier, the integrand is purely real and is being integral on a real interval. I noted that there was an interesting symmetry in the exponentials, but haven't found a way to use that.



I thought that it was possible that a mistake related to phase occurred, but I wouldn't be sure where.



Any ideas?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
    $endgroup$
    – Winther
    Dec 22 '18 at 21:49








  • 1




    $begingroup$
    For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
    $endgroup$
    – Winther
    Dec 22 '18 at 22:06
















4












$begingroup$


I have been trying to calculate the following integral using the Residue Theorem but end up with an answer that seems to contain an imaginary part while the integral should be purely real:
$$I_1=int_0^infty frac{cos(a x)}{x^4+b^4},dx$$
In order to calculate this integral, I consider the contour integral
$$I_2=int_C frac{e^{iaz}}{z^4+b^4},dz$$
Where $C$ is the semi-circle of radius $R$ in the upper half plane parameterized counter-clockwise. As $Rtoinfty$, the semi-circular integral vanishes, and a symmetry argument shows that $I_1=I_2/2$.



Now, we can use the Residue Theorem to compute $I_2$, which has poles at $z=be^{ipi/4}$ and $z=be^{3ipi/4}$. After some computations, I obtain
begin{align}I_1&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}+e^{-pi ab /4}e^{-3ipi/4} right)\
&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}-ie^{-pi ab /4}e^{-ipi/4} right) \
&=frac{pi i}{4b^3} e^{-ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
&=frac{pi}{4b^3} e^{ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
end{align}

Now, it seems that unless $a=0$ ( and $b=0$, but at that point the integral is undefined), there will always be a non-zero imaginary part, which shouldn't occur because, as I said earlier, the integrand is purely real and is being integral on a real interval. I noted that there was an interesting symmetry in the exponentials, but haven't found a way to use that.



I thought that it was possible that a mistake related to phase occurred, but I wouldn't be sure where.



Any ideas?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
    $endgroup$
    – Winther
    Dec 22 '18 at 21:49








  • 1




    $begingroup$
    For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
    $endgroup$
    – Winther
    Dec 22 '18 at 22:06














4












4








4


2



$begingroup$


I have been trying to calculate the following integral using the Residue Theorem but end up with an answer that seems to contain an imaginary part while the integral should be purely real:
$$I_1=int_0^infty frac{cos(a x)}{x^4+b^4},dx$$
In order to calculate this integral, I consider the contour integral
$$I_2=int_C frac{e^{iaz}}{z^4+b^4},dz$$
Where $C$ is the semi-circle of radius $R$ in the upper half plane parameterized counter-clockwise. As $Rtoinfty$, the semi-circular integral vanishes, and a symmetry argument shows that $I_1=I_2/2$.



Now, we can use the Residue Theorem to compute $I_2$, which has poles at $z=be^{ipi/4}$ and $z=be^{3ipi/4}$. After some computations, I obtain
begin{align}I_1&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}+e^{-pi ab /4}e^{-3ipi/4} right)\
&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}-ie^{-pi ab /4}e^{-ipi/4} right) \
&=frac{pi i}{4b^3} e^{-ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
&=frac{pi}{4b^3} e^{ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
end{align}

Now, it seems that unless $a=0$ ( and $b=0$, but at that point the integral is undefined), there will always be a non-zero imaginary part, which shouldn't occur because, as I said earlier, the integrand is purely real and is being integral on a real interval. I noted that there was an interesting symmetry in the exponentials, but haven't found a way to use that.



I thought that it was possible that a mistake related to phase occurred, but I wouldn't be sure where.



Any ideas?










share|cite|improve this question









$endgroup$




I have been trying to calculate the following integral using the Residue Theorem but end up with an answer that seems to contain an imaginary part while the integral should be purely real:
$$I_1=int_0^infty frac{cos(a x)}{x^4+b^4},dx$$
In order to calculate this integral, I consider the contour integral
$$I_2=int_C frac{e^{iaz}}{z^4+b^4},dz$$
Where $C$ is the semi-circle of radius $R$ in the upper half plane parameterized counter-clockwise. As $Rtoinfty$, the semi-circular integral vanishes, and a symmetry argument shows that $I_1=I_2/2$.



Now, we can use the Residue Theorem to compute $I_2$, which has poles at $z=be^{ipi/4}$ and $z=be^{3ipi/4}$. After some computations, I obtain
begin{align}I_1&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}+e^{-pi ab /4}e^{-3ipi/4} right)\
&=frac{pi i}{4b^3} left( e^{-3pi ab /4}e^{-ipi/4}-ie^{-pi ab /4}e^{-ipi/4} right) \
&=frac{pi i}{4b^3} e^{-ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
&=frac{pi}{4b^3} e^{ipi/4}left( e^{-3pi ab /4}-ie^{-pi ab /4} right) \
end{align}

Now, it seems that unless $a=0$ ( and $b=0$, but at that point the integral is undefined), there will always be a non-zero imaginary part, which shouldn't occur because, as I said earlier, the integrand is purely real and is being integral on a real interval. I noted that there was an interesting symmetry in the exponentials, but haven't found a way to use that.



I thought that it was possible that a mistake related to phase occurred, but I wouldn't be sure where.



Any ideas?







calculus integration contour-integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 '18 at 21:14









ZacharyZachary

2,3751214




2,3751214












  • $begingroup$
    The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
    $endgroup$
    – Winther
    Dec 22 '18 at 21:49








  • 1




    $begingroup$
    For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
    $endgroup$
    – Winther
    Dec 22 '18 at 22:06


















  • $begingroup$
    The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
    $endgroup$
    – Winther
    Dec 22 '18 at 21:49








  • 1




    $begingroup$
    For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
    $endgroup$
    – Winther
    Dec 22 '18 at 22:06
















$begingroup$
The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
$endgroup$
– Winther
Dec 22 '18 at 21:49






$begingroup$
The problem seems to be in the computations of the residues (I find $-frac{(1+i)e^{-frac{(1-i) a b}{sqrt{2}}}}{4sqrt{2}b^3}$ for the first residue). btw to simplify the analysis note that by a substitution then if the integral is $f(a,b)$ then we have $f(a,b) = frac{1}{|b|^3}f(|ab|,1)$ so you can take $b=1$ and then put the $b$-dependency back in the end using this scaling.
$endgroup$
– Winther
Dec 22 '18 at 21:49






1




1




$begingroup$
For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
$endgroup$
– Winther
Dec 22 '18 at 22:06




$begingroup$
For reference: if you compute the residues correctly and simplify you should end up with $frac{pi e^{-frac{| a b| }{sqrt{2}}} sinleft(frac{|ab|}{sqrt{2}}+frac{pi }{4}right)}{2|b|^3}$
$endgroup$
– Winther
Dec 22 '18 at 22:06










2 Answers
2






active

oldest

votes


















2












$begingroup$

Alright, I found an entirely real answer which agrees with Winther's comment. I originally made a mistake when calculating my residues. Here we go:



begin{align}
I_2&= 2pi i sum text{Res}_{text{Im(}z)ge 0 } ,f(z) \
&= frac{2pi i}{4b^3}left(frac{e^{iabe^{3pi i/4}}}{e^{ipi /4}} + frac{e^{iabe^{pi i/4}}}{e^{3ipi /4}} right) \
&= frac{2pi i}{4b^3}left(e^{iableft(-frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{ipi}{4}} + e^{iableft(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{3ipi}{4}} right) \
&=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}+ e^{ileft[frac{ab}{sqrt{2}} -frac{3pi}{4}right]} right) \
&=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}- e^{ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]} right) \
&=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( -2 isin left(frac{ab}{sqrt{2}} +frac{pi}{4}right) right) \
&= frac{pi}{b^3} e^{-frac{ab}{sqrt{2}}}sin left(frac{ab}{sqrt{2}} +frac{pi}{4}right)
end{align}

using $I_1=I_2/2$ and demanding that $I(a)=I(-a)$ and $I(b)=I(-b)$, we finally obtain



$$I_1=frac{pi}{2b^3} e^{-frac{|ab|}{sqrt{2}}}sin left(frac{|ab|}{sqrt{2}} +frac{pi}{4}right)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
    $endgroup$
    – Poujh
    Dec 22 '18 at 23:26












  • $begingroup$
    Yes, I just fixed it. Thank you for the comment.
    $endgroup$
    – Zachary
    Dec 23 '18 at 3:02



















1












$begingroup$

Not really an answer, just making things as compact as possible



First of all note that $$e^{ipi/4}=frac{1+i}{sqrt2}$$
Hence
$$I_1=frac{pi(1+i)}{4b^3sqrt2}bigg(frac1{e^{3abpi/4}}-frac{i}{e^{abpi/4}}bigg)$$
Multiplying the RHS by $frac{exp(abpi/4)}{exp(abpi/4)}$,
$$I_1=frac{pi(1+i)}{4b^3e^{abpi/4}sqrt2}bigg(e^{frac{abpi}4(1-3)}-ie^{frac{abpi}4(1-1)}bigg)$$
$$I_1=frac{pi(1+i)}{b^3e^{abpi/4}2^{5/2}}(e^{-abpi/2}-i)$$
Now let $$p=fracpi{b^3e^{abpi/4}2^{5/2}}\ q=e^{-abpi/2}$$
So
$$I_1=p(1+i)(q-i)$$
Thus
$$I_1=p(1+q)+ip(1-q)$$
So now you have a distinct real and imaginary part. I guess you have to make more assumptions about the restrictions on $a$ and $b$ if you want $text{Im}I_1=0$. Either that or you made a mistake.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049850%2fhow-to-simplify-the-result-i-obtain-for-the-following-integral-int-0-infty%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Alright, I found an entirely real answer which agrees with Winther's comment. I originally made a mistake when calculating my residues. Here we go:



    begin{align}
    I_2&= 2pi i sum text{Res}_{text{Im(}z)ge 0 } ,f(z) \
    &= frac{2pi i}{4b^3}left(frac{e^{iabe^{3pi i/4}}}{e^{ipi /4}} + frac{e^{iabe^{pi i/4}}}{e^{3ipi /4}} right) \
    &= frac{2pi i}{4b^3}left(e^{iableft(-frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{ipi}{4}} + e^{iableft(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{3ipi}{4}} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}+ e^{ileft[frac{ab}{sqrt{2}} -frac{3pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}- e^{ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( -2 isin left(frac{ab}{sqrt{2}} +frac{pi}{4}right) right) \
    &= frac{pi}{b^3} e^{-frac{ab}{sqrt{2}}}sin left(frac{ab}{sqrt{2}} +frac{pi}{4}right)
    end{align}

    using $I_1=I_2/2$ and demanding that $I(a)=I(-a)$ and $I(b)=I(-b)$, we finally obtain



    $$I_1=frac{pi}{2b^3} e^{-frac{|ab|}{sqrt{2}}}sin left(frac{|ab|}{sqrt{2}} +frac{pi}{4}right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
      $endgroup$
      – Poujh
      Dec 22 '18 at 23:26












    • $begingroup$
      Yes, I just fixed it. Thank you for the comment.
      $endgroup$
      – Zachary
      Dec 23 '18 at 3:02
















    2












    $begingroup$

    Alright, I found an entirely real answer which agrees with Winther's comment. I originally made a mistake when calculating my residues. Here we go:



    begin{align}
    I_2&= 2pi i sum text{Res}_{text{Im(}z)ge 0 } ,f(z) \
    &= frac{2pi i}{4b^3}left(frac{e^{iabe^{3pi i/4}}}{e^{ipi /4}} + frac{e^{iabe^{pi i/4}}}{e^{3ipi /4}} right) \
    &= frac{2pi i}{4b^3}left(e^{iableft(-frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{ipi}{4}} + e^{iableft(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{3ipi}{4}} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}+ e^{ileft[frac{ab}{sqrt{2}} -frac{3pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}- e^{ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( -2 isin left(frac{ab}{sqrt{2}} +frac{pi}{4}right) right) \
    &= frac{pi}{b^3} e^{-frac{ab}{sqrt{2}}}sin left(frac{ab}{sqrt{2}} +frac{pi}{4}right)
    end{align}

    using $I_1=I_2/2$ and demanding that $I(a)=I(-a)$ and $I(b)=I(-b)$, we finally obtain



    $$I_1=frac{pi}{2b^3} e^{-frac{|ab|}{sqrt{2}}}sin left(frac{|ab|}{sqrt{2}} +frac{pi}{4}right)$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
      $endgroup$
      – Poujh
      Dec 22 '18 at 23:26












    • $begingroup$
      Yes, I just fixed it. Thank you for the comment.
      $endgroup$
      – Zachary
      Dec 23 '18 at 3:02














    2












    2








    2





    $begingroup$

    Alright, I found an entirely real answer which agrees with Winther's comment. I originally made a mistake when calculating my residues. Here we go:



    begin{align}
    I_2&= 2pi i sum text{Res}_{text{Im(}z)ge 0 } ,f(z) \
    &= frac{2pi i}{4b^3}left(frac{e^{iabe^{3pi i/4}}}{e^{ipi /4}} + frac{e^{iabe^{pi i/4}}}{e^{3ipi /4}} right) \
    &= frac{2pi i}{4b^3}left(e^{iableft(-frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{ipi}{4}} + e^{iableft(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{3ipi}{4}} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}+ e^{ileft[frac{ab}{sqrt{2}} -frac{3pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}- e^{ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( -2 isin left(frac{ab}{sqrt{2}} +frac{pi}{4}right) right) \
    &= frac{pi}{b^3} e^{-frac{ab}{sqrt{2}}}sin left(frac{ab}{sqrt{2}} +frac{pi}{4}right)
    end{align}

    using $I_1=I_2/2$ and demanding that $I(a)=I(-a)$ and $I(b)=I(-b)$, we finally obtain



    $$I_1=frac{pi}{2b^3} e^{-frac{|ab|}{sqrt{2}}}sin left(frac{|ab|}{sqrt{2}} +frac{pi}{4}right)$$






    share|cite|improve this answer











    $endgroup$



    Alright, I found an entirely real answer which agrees with Winther's comment. I originally made a mistake when calculating my residues. Here we go:



    begin{align}
    I_2&= 2pi i sum text{Res}_{text{Im(}z)ge 0 } ,f(z) \
    &= frac{2pi i}{4b^3}left(frac{e^{iabe^{3pi i/4}}}{e^{ipi /4}} + frac{e^{iabe^{pi i/4}}}{e^{3ipi /4}} right) \
    &= frac{2pi i}{4b^3}left(e^{iableft(-frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{ipi}{4}} + e^{iableft(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2}right)-frac{3ipi}{4}} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}+ e^{ileft[frac{ab}{sqrt{2}} -frac{3pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( e^{-ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]}- e^{ileft[frac{ab}{sqrt{2}} +frac{pi}{4}right]} right) \
    &=frac{2pi i}{4b^3}e^{-frac{ab}{sqrt{2}}} left( -2 isin left(frac{ab}{sqrt{2}} +frac{pi}{4}right) right) \
    &= frac{pi}{b^3} e^{-frac{ab}{sqrt{2}}}sin left(frac{ab}{sqrt{2}} +frac{pi}{4}right)
    end{align}

    using $I_1=I_2/2$ and demanding that $I(a)=I(-a)$ and $I(b)=I(-b)$, we finally obtain



    $$I_1=frac{pi}{2b^3} e^{-frac{|ab|}{sqrt{2}}}sin left(frac{|ab|}{sqrt{2}} +frac{pi}{4}right)$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 3:02

























    answered Dec 22 '18 at 22:47









    ZacharyZachary

    2,3751214




    2,3751214












    • $begingroup$
      I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
      $endgroup$
      – Poujh
      Dec 22 '18 at 23:26












    • $begingroup$
      Yes, I just fixed it. Thank you for the comment.
      $endgroup$
      – Zachary
      Dec 23 '18 at 3:02


















    • $begingroup$
      I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
      $endgroup$
      – Poujh
      Dec 22 '18 at 23:26












    • $begingroup$
      Yes, I just fixed it. Thank you for the comment.
      $endgroup$
      – Zachary
      Dec 23 '18 at 3:02
















    $begingroup$
    I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
    $endgroup$
    – Poujh
    Dec 22 '18 at 23:26






    $begingroup$
    I think your forgot the b in the exponent on row 2 and 3 (as a typo I mean because you have it again on row 4)
    $endgroup$
    – Poujh
    Dec 22 '18 at 23:26














    $begingroup$
    Yes, I just fixed it. Thank you for the comment.
    $endgroup$
    – Zachary
    Dec 23 '18 at 3:02




    $begingroup$
    Yes, I just fixed it. Thank you for the comment.
    $endgroup$
    – Zachary
    Dec 23 '18 at 3:02











    1












    $begingroup$

    Not really an answer, just making things as compact as possible



    First of all note that $$e^{ipi/4}=frac{1+i}{sqrt2}$$
    Hence
    $$I_1=frac{pi(1+i)}{4b^3sqrt2}bigg(frac1{e^{3abpi/4}}-frac{i}{e^{abpi/4}}bigg)$$
    Multiplying the RHS by $frac{exp(abpi/4)}{exp(abpi/4)}$,
    $$I_1=frac{pi(1+i)}{4b^3e^{abpi/4}sqrt2}bigg(e^{frac{abpi}4(1-3)}-ie^{frac{abpi}4(1-1)}bigg)$$
    $$I_1=frac{pi(1+i)}{b^3e^{abpi/4}2^{5/2}}(e^{-abpi/2}-i)$$
    Now let $$p=fracpi{b^3e^{abpi/4}2^{5/2}}\ q=e^{-abpi/2}$$
    So
    $$I_1=p(1+i)(q-i)$$
    Thus
    $$I_1=p(1+q)+ip(1-q)$$
    So now you have a distinct real and imaginary part. I guess you have to make more assumptions about the restrictions on $a$ and $b$ if you want $text{Im}I_1=0$. Either that or you made a mistake.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Not really an answer, just making things as compact as possible



      First of all note that $$e^{ipi/4}=frac{1+i}{sqrt2}$$
      Hence
      $$I_1=frac{pi(1+i)}{4b^3sqrt2}bigg(frac1{e^{3abpi/4}}-frac{i}{e^{abpi/4}}bigg)$$
      Multiplying the RHS by $frac{exp(abpi/4)}{exp(abpi/4)}$,
      $$I_1=frac{pi(1+i)}{4b^3e^{abpi/4}sqrt2}bigg(e^{frac{abpi}4(1-3)}-ie^{frac{abpi}4(1-1)}bigg)$$
      $$I_1=frac{pi(1+i)}{b^3e^{abpi/4}2^{5/2}}(e^{-abpi/2}-i)$$
      Now let $$p=fracpi{b^3e^{abpi/4}2^{5/2}}\ q=e^{-abpi/2}$$
      So
      $$I_1=p(1+i)(q-i)$$
      Thus
      $$I_1=p(1+q)+ip(1-q)$$
      So now you have a distinct real and imaginary part. I guess you have to make more assumptions about the restrictions on $a$ and $b$ if you want $text{Im}I_1=0$. Either that or you made a mistake.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Not really an answer, just making things as compact as possible



        First of all note that $$e^{ipi/4}=frac{1+i}{sqrt2}$$
        Hence
        $$I_1=frac{pi(1+i)}{4b^3sqrt2}bigg(frac1{e^{3abpi/4}}-frac{i}{e^{abpi/4}}bigg)$$
        Multiplying the RHS by $frac{exp(abpi/4)}{exp(abpi/4)}$,
        $$I_1=frac{pi(1+i)}{4b^3e^{abpi/4}sqrt2}bigg(e^{frac{abpi}4(1-3)}-ie^{frac{abpi}4(1-1)}bigg)$$
        $$I_1=frac{pi(1+i)}{b^3e^{abpi/4}2^{5/2}}(e^{-abpi/2}-i)$$
        Now let $$p=fracpi{b^3e^{abpi/4}2^{5/2}}\ q=e^{-abpi/2}$$
        So
        $$I_1=p(1+i)(q-i)$$
        Thus
        $$I_1=p(1+q)+ip(1-q)$$
        So now you have a distinct real and imaginary part. I guess you have to make more assumptions about the restrictions on $a$ and $b$ if you want $text{Im}I_1=0$. Either that or you made a mistake.






        share|cite|improve this answer









        $endgroup$



        Not really an answer, just making things as compact as possible



        First of all note that $$e^{ipi/4}=frac{1+i}{sqrt2}$$
        Hence
        $$I_1=frac{pi(1+i)}{4b^3sqrt2}bigg(frac1{e^{3abpi/4}}-frac{i}{e^{abpi/4}}bigg)$$
        Multiplying the RHS by $frac{exp(abpi/4)}{exp(abpi/4)}$,
        $$I_1=frac{pi(1+i)}{4b^3e^{abpi/4}sqrt2}bigg(e^{frac{abpi}4(1-3)}-ie^{frac{abpi}4(1-1)}bigg)$$
        $$I_1=frac{pi(1+i)}{b^3e^{abpi/4}2^{5/2}}(e^{-abpi/2}-i)$$
        Now let $$p=fracpi{b^3e^{abpi/4}2^{5/2}}\ q=e^{-abpi/2}$$
        So
        $$I_1=p(1+i)(q-i)$$
        Thus
        $$I_1=p(1+q)+ip(1-q)$$
        So now you have a distinct real and imaginary part. I guess you have to make more assumptions about the restrictions on $a$ and $b$ if you want $text{Im}I_1=0$. Either that or you made a mistake.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 21:53









        clathratusclathratus

        4,745337




        4,745337






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3049850%2fhow-to-simplify-the-result-i-obtain-for-the-following-integral-int-0-infty%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Wiesbaden

            Marschland

            Dieringhausen