Integration problem solving without contour integration
Can the following question be solved without using contour integration.
$F:(0,infty)times (0,infty)to Bbb R$ be given by
$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(alpha x)}{x^4+beta^4},dx$
Show that
$frac{F(alpha,beta)}{F(beta,alpha)}=frac{alpha^3}{beta^3}$ as long as there is no positive integer n such that $alpha=frac{(4n-1)pisqrt{2}}{4beta}$
integration improper-integrals self-learning contour-integration
asked Dec 22 at 12:56
Dhamnekar Winod
385414
|
show 2 more comments
Can the following question be solved without using contour integration.
$F:(0,infty)times (0,infty)to Bbb R$ be given by
$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(alpha x)}{x^4+beta^4},dx$
Show that
$frac{F(alpha,beta)}{F(beta,alpha)}=frac{alpha^3}{beta^3}$ as long as there is no positive integer n such that $alpha=frac{(4n-1)pisqrt{2}}{4beta}$
integration improper-integrals self-learning contour-integration
asked Dec 22 at 12:56
Dhamnekar Winod
385414
I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21
|
show 2 more comments
Can the following question be solved without using contour integration.
$F:(0,infty)times (0,infty)to Bbb R$ be given by
$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(alpha x)}{x^4+beta^4},dx$
Show that
$frac{F(alpha,beta)}{F(beta,alpha)}=frac{alpha^3}{beta^3}$ as long as there is no positive integer n such that $alpha=frac{(4n-1)pisqrt{2}}{4beta}$
integration improper-integrals self-learning contour-integration
asked Dec 22 at 12:56
Dhamnekar Winod
385414
Can the following question be solved without using contour integration.
$F:(0,infty)times (0,infty)to Bbb R$ be given by
$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(alpha x)}{x^4+beta^4},dx$
Show that
$frac{F(alpha,beta)}{F(beta,alpha)}=frac{alpha^3}{beta^3}$ as long as there is no positive integer n such that $alpha=frac{(4n-1)pisqrt{2}}{4beta}$
integration improper-integrals self-learning contour-integration
integration improper-integrals self-learning contour-integration
asked Dec 22 at 12:56
Dhamnekar Winod
385414
asked Dec 22 at 12:56
Dhamnekar Winod
385414
asked Dec 22 at 12:56
Dhamnekar Winod
385414
asked Dec 22 at 12:56
Dhamnekar Winod
385414
asked Dec 22 at 12:56
Dhamnekar Winod
385414
385414
I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21
|
show 2 more comments
I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21
I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21
|
show 2 more comments
2 Answers
2
active
oldest
votes
Switching to integration parameter $t=frac{alpha}{beta}x$ gives $$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(beta t)}{(frac{beta}{alpha}t)^4+beta^4},frac{beta}{alpha},dt=frac{alpha^3}{beta^3}F(beta,alpha)$$
answered Dec 22 at 13:55
random
47126
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
add a comment |
To answer your question concerning why the quotient
$$frac{F(alpha, beta)}{F(beta, alpha)} = frac{alpha^3}{beta^3},$$
is only defined provided
$$alpha neq frac{(4n - 1) pi sqrt{2}}{4b},$$
where $n in mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.
Using the Residue Theorem a value for $F(alpha, beta)$ can be found. It is given by (for a proof of this, see here)
$$F (alpha, beta) = frac{pi}{2 alpha^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
Similarly,
$$F (beta, alpha) = frac{pi}{2 beta^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
If the required quotient is to be defined, clearly $F (beta, alpha) neq 0$. Thus the sine term appearing in the expression for $F(beta, alpha)$ must be non-zero. Thus
$$frac{alpha beta}{sqrt{2}} + frac{pi}{4} neq n pi quad text{or} quad alpha neq frac{(4n - 1) pi sqrt{2}}{4 beta},$$
where $n in mathbb{N}$, as required.
answered Dec 23 at 9:40
omegadot
4,5842725
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Switching to integration parameter $t=frac{alpha}{beta}x$ gives $$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(beta t)}{(frac{beta}{alpha}t)^4+beta^4},frac{beta}{alpha},dt=frac{alpha^3}{beta^3}F(beta,alpha)$$
answered Dec 22 at 13:55
random
47126
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
add a comment |
Switching to integration parameter $t=frac{alpha}{beta}x$ gives $$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(beta t)}{(frac{beta}{alpha}t)^4+beta^4},frac{beta}{alpha},dt=frac{alpha^3}{beta^3}F(beta,alpha)$$
answered Dec 22 at 13:55
random
47126
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
add a comment |
Switching to integration parameter $t=frac{alpha}{beta}x$ gives $$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(beta t)}{(frac{beta}{alpha}t)^4+beta^4},frac{beta}{alpha},dt=frac{alpha^3}{beta^3}F(beta,alpha)$$
answered Dec 22 at 13:55
random
47126
Switching to integration parameter $t=frac{alpha}{beta}x$ gives $$F(alpha,beta)=displaystyleint_0^inftyfrac{cos(beta t)}{(frac{beta}{alpha}t)^4+beta^4},frac{beta}{alpha},dt=frac{alpha^3}{beta^3}F(beta,alpha)$$
answered Dec 22 at 13:55
random
47126
edited Dec 22 at 14:00
answered Dec 22 at 13:55
random
47126
answered Dec 22 at 13:55
random
47126
answered Dec 22 at 13:55
random
47126
47126
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
add a comment |
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
1
1
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
,what about the condition of positive integer n?
– Dhamnekar Winod
Dec 22 at 16:03
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
As mentioned by Zachary, for such a special $bf{n}$ one has $F(alpha,beta)=F(beta,alpha)=0$, so their quotient is undefined. Rewriting the equation as ${beta}^3F(alpha,beta)={alpha}^3F(beta,alpha)$ avoids that problem.
– random
Dec 23 at 1:26
add a comment |
To answer your question concerning why the quotient
$$frac{F(alpha, beta)}{F(beta, alpha)} = frac{alpha^3}{beta^3},$$
is only defined provided
$$alpha neq frac{(4n - 1) pi sqrt{2}}{4b},$$
where $n in mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.
Using the Residue Theorem a value for $F(alpha, beta)$ can be found. It is given by (for a proof of this, see here)
$$F (alpha, beta) = frac{pi}{2 alpha^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
Similarly,
$$F (beta, alpha) = frac{pi}{2 beta^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
If the required quotient is to be defined, clearly $F (beta, alpha) neq 0$. Thus the sine term appearing in the expression for $F(beta, alpha)$ must be non-zero. Thus
$$frac{alpha beta}{sqrt{2}} + frac{pi}{4} neq n pi quad text{or} quad alpha neq frac{(4n - 1) pi sqrt{2}}{4 beta},$$
where $n in mathbb{N}$, as required.
answered Dec 23 at 9:40
omegadot
4,5842725
add a comment |
To answer your question concerning why the quotient
$$frac{F(alpha, beta)}{F(beta, alpha)} = frac{alpha^3}{beta^3},$$
is only defined provided
$$alpha neq frac{(4n - 1) pi sqrt{2}}{4b},$$
where $n in mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.
Using the Residue Theorem a value for $F(alpha, beta)$ can be found. It is given by (for a proof of this, see here)
$$F (alpha, beta) = frac{pi}{2 alpha^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
Similarly,
$$F (beta, alpha) = frac{pi}{2 beta^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
If the required quotient is to be defined, clearly $F (beta, alpha) neq 0$. Thus the sine term appearing in the expression for $F(beta, alpha)$ must be non-zero. Thus
$$frac{alpha beta}{sqrt{2}} + frac{pi}{4} neq n pi quad text{or} quad alpha neq frac{(4n - 1) pi sqrt{2}}{4 beta},$$
where $n in mathbb{N}$, as required.
answered Dec 23 at 9:40
omegadot
4,5842725
add a comment |
To answer your question concerning why the quotient
$$frac{F(alpha, beta)}{F(beta, alpha)} = frac{alpha^3}{beta^3},$$
is only defined provided
$$alpha neq frac{(4n - 1) pi sqrt{2}}{4b},$$
where $n in mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.
Using the Residue Theorem a value for $F(alpha, beta)$ can be found. It is given by (for a proof of this, see here)
$$F (alpha, beta) = frac{pi}{2 alpha^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
Similarly,
$$F (beta, alpha) = frac{pi}{2 beta^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
If the required quotient is to be defined, clearly $F (beta, alpha) neq 0$. Thus the sine term appearing in the expression for $F(beta, alpha)$ must be non-zero. Thus
$$frac{alpha beta}{sqrt{2}} + frac{pi}{4} neq n pi quad text{or} quad alpha neq frac{(4n - 1) pi sqrt{2}}{4 beta},$$
where $n in mathbb{N}$, as required.
answered Dec 23 at 9:40
omegadot
4,5842725
To answer your question concerning why the quotient
$$frac{F(alpha, beta)}{F(beta, alpha)} = frac{alpha^3}{beta^3},$$
is only defined provided
$$alpha neq frac{(4n - 1) pi sqrt{2}}{4b},$$
where $n in mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.
Using the Residue Theorem a value for $F(alpha, beta)$ can be found. It is given by (for a proof of this, see here)
$$F (alpha, beta) = frac{pi}{2 alpha^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
Similarly,
$$F (beta, alpha) = frac{pi}{2 beta^3} exp left (-frac{alpha beta}{sqrt{2}} right ) sin left (frac{alpha beta}{sqrt{2}} + frac{pi}{4} right ), qquad alpha, beta > 0.$$
If the required quotient is to be defined, clearly $F (beta, alpha) neq 0$. Thus the sine term appearing in the expression for $F(beta, alpha)$ must be non-zero. Thus
$$frac{alpha beta}{sqrt{2}} + frac{pi}{4} neq n pi quad text{or} quad alpha neq frac{(4n - 1) pi sqrt{2}}{4 beta},$$
where $n in mathbb{N}$, as required.
answered Dec 23 at 9:40
omegadot
4,5842725
answered Dec 23 at 9:40
omegadot
4,5842725
answered Dec 23 at 9:40
omegadot
4,5842725
answered Dec 23 at 9:40
omegadot
4,5842725
4,5842725
add a comment |
add a comment |
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I gave it a few tries, but I am not able to solve it by Feynman integration. Integrating by power series also seems unlikely to succeed. And obviously finding an anti-derivative is completely hopeless as it is not elementary.
– SmileyCraft
Dec 22 at 13:45
@SmileyCraft. In fact, there is an explicit formula for the integral.
– Claude Leibovici
Dec 22 at 15:11
@ClaudeLeibovici Yeah, but do you know of a way of finding it without contour integration? And all I said was there is no explicit formula for the antiderivative of the integrand.
– SmileyCraft
Dec 22 at 15:15
@SmileyCraft. A CAS made it for me (the integral) ! The antiderivative is quite awful but doable in terms of sine and cosine integrals after partial fracion decomposition.
– Claude Leibovici
Dec 22 at 15:18
@ClaudeLeibovici I meant that it is not elementary. But indeed there is an antiderivative, but there always is one for continuous functions. By the way, what is a CAS?
– SmileyCraft
Dec 22 at 15:21