Mixing
Integral of the form $e^i K sqrtx^2+a^2/sqrtx^2+a^2$
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I am trying to solve integrals of the following form $$intfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x,$$
where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?
Substituting the square root doesn't seem to help since for $p = sqrtx^2+a^2$ we have $frac rm dprm dx = frac x p = fracsqrtp^2-a^2p$, leading to $$intfrac rm e^imath K psqrtp^2-a^2 rm d p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.
Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.
Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?
edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$int_-infty^inftyfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x = -pi(Y_0(a K) + jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.
integration
asked Jul 18 at 11:41
Florian
1,2741617
 |Â
show 1 more comment
up vote
0
down vote
favorite
I am trying to solve integrals of the following form $$intfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x,$$
where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?
Substituting the square root doesn't seem to help since for $p = sqrtx^2+a^2$ we have $frac rm dprm dx = frac x p = fracsqrtp^2-a^2p$, leading to $$intfrac rm e^imath K psqrtp^2-a^2 rm d p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.
Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.
Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?
edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$int_-infty^inftyfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x = -pi(Y_0(a K) + jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.
integration
asked Jul 18 at 11:41
Florian
1,2741617
Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve integrals of the following form $$intfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x,$$
where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?
Substituting the square root doesn't seem to help since for $p = sqrtx^2+a^2$ we have $frac rm dprm dx = frac x p = fracsqrtp^2-a^2p$, leading to $$intfrac rm e^imath K psqrtp^2-a^2 rm d p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.
Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.
Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?
edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$int_-infty^inftyfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x = -pi(Y_0(a K) + jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.
integration
asked Jul 18 at 11:41
Florian
1,2741617
I am trying to solve integrals of the following form $$intfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x,$$
where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?
Substituting the square root doesn't seem to help since for $p = sqrtx^2+a^2$ we have $frac rm dprm dx = frac x p = fracsqrtp^2-a^2p$, leading to $$intfrac rm e^imath K psqrtp^2-a^2 rm d p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.
Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.
Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?
edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$int_-infty^inftyfrac rm e^imath K sqrtx^2+a^2sqrtx^2+a^2 rm d x = -pi(Y_0(a K) + jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.
integration
asked Jul 18 at 11:41
Florian
1,2741617
edited Jul 18 at 15:30
asked Jul 18 at 11:41
Florian
1,2741617
asked Jul 18 at 11:41
Florian
1,2741617
asked Jul 18 at 11:41
Florian
1,2741617
1,2741617
Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22
 |Â
show 1 more comment
Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22
Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Changing $x=asinh t$ in the integral,
beginequation
I=int_beta^gammae^ikacosh t,dt
endequation
where $beta=sinh^-1b/a$ and $gamma=sinh^-1(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<sigma<pi$
beginequation
H_0^(1)(ka,w)=frac2ipiint_w^infty+isigmae^ikacosh t,dt
endequation
Then
beginequation
I=frac2ipileft[ H_0^(1)(ka,beta)- H_0^(1)(ka,gamma) right]
endequation
It can also be expressed as
beginequation
I= K_0(-ika,beta)- K_0(-ika,gamma)
endequation
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
answered Jul 18 at 20:15
Paul Enta
3,3271925
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
add a comment |Â
up vote
0
down vote
Consider function:
$$y=sqrtx^2+a^2 ln e^sqrt x^2 +a^2$$
$$y'=fracxsqrt x^2+a^2.(e^ sqrtx^2+a^2)(1+sqrtx^2+a^2)$$
Now compare this with the relation in question we get:
$$k=-[frac ln x(1+sqrtx^2+a^2)sqrt x^2+a^2+1]i$$
This is condition for initial relation to be integrable.
answered Jul 18 at 17:35
sirous
761511
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Changing $x=asinh t$ in the integral,
beginequation
I=int_beta^gammae^ikacosh t,dt
endequation
where $beta=sinh^-1b/a$ and $gamma=sinh^-1(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<sigma<pi$
beginequation
H_0^(1)(ka,w)=frac2ipiint_w^infty+isigmae^ikacosh t,dt
endequation
Then
beginequation
I=frac2ipileft[ H_0^(1)(ka,beta)- H_0^(1)(ka,gamma) right]
endequation
It can also be expressed as
beginequation
I= K_0(-ika,beta)- K_0(-ika,gamma)
endequation
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
answered Jul 18 at 20:15
Paul Enta
3,3271925
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
add a comment |Â
up vote
1
down vote
accepted
Changing $x=asinh t$ in the integral,
beginequation
I=int_beta^gammae^ikacosh t,dt
endequation
where $beta=sinh^-1b/a$ and $gamma=sinh^-1(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<sigma<pi$
beginequation
H_0^(1)(ka,w)=frac2ipiint_w^infty+isigmae^ikacosh t,dt
endequation
Then
beginequation
I=frac2ipileft[ H_0^(1)(ka,beta)- H_0^(1)(ka,gamma) right]
endequation
It can also be expressed as
beginequation
I= K_0(-ika,beta)- K_0(-ika,gamma)
endequation
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
answered Jul 18 at 20:15
Paul Enta
3,3271925
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Changing $x=asinh t$ in the integral,
beginequation
I=int_beta^gammae^ikacosh t,dt
endequation
where $beta=sinh^-1b/a$ and $gamma=sinh^-1(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<sigma<pi$
beginequation
H_0^(1)(ka,w)=frac2ipiint_w^infty+isigmae^ikacosh t,dt
endequation
Then
beginequation
I=frac2ipileft[ H_0^(1)(ka,beta)- H_0^(1)(ka,gamma) right]
endequation
It can also be expressed as
beginequation
I= K_0(-ika,beta)- K_0(-ika,gamma)
endequation
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
answered Jul 18 at 20:15
Paul Enta
3,3271925
Changing $x=asinh t$ in the integral,
beginequation
I=int_beta^gammae^ikacosh t,dt
endequation
where $beta=sinh^-1b/a$ and $gamma=sinh^-1(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<sigma<pi$
beginequation
H_0^(1)(ka,w)=frac2ipiint_w^infty+isigmae^ikacosh t,dt
endequation
Then
beginequation
I=frac2ipileft[ H_0^(1)(ka,beta)- H_0^(1)(ka,gamma) right]
endequation
It can also be expressed as
beginequation
I= K_0(-ika,beta)- K_0(-ika,gamma)
endequation
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
answered Jul 18 at 20:15
Paul Enta
3,3271925
edited Jul 19 at 9:45
answered Jul 18 at 20:15
Paul Enta
3,3271925
answered Jul 18 at 20:15
Paul Enta
3,3271925
answered Jul 18 at 20:15
Paul Enta
3,3271925
3,3271925
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
add a comment |Â
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
Ah, what a wonderful substitution. Thanks so much, that is it! As a minor thing, it seems I would have to change the integration limits since for $x$ to run from $b$ to $b+c$, I need to let $t$ run from something like $rm arsinh(b/a)$. But that's a minor thing. Thank you again!
– Florian
Jul 19 at 8:40
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
What still confuses me is that the integral needs to run on a line with $rm Im(t)>0$ (that's how I interpret the $i sigma$ thing?). After the substitution, the integral is strictly on the real line though. What's up with that?
– Florian
Jul 19 at 8:44
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
Sorry for the mistake on the integration limits, corrected now.
– Paul Enta
Jul 19 at 9:46
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
This can be viewed as writing that the integral on a closed contour of a holomorphic function in the complex plane is zero (residue theorem). This contour is made of 3 components 1) from $beta$ to $gamma$ on the real axis, 2) from $gamma$ to $infty +isigma$ and 3) from $infty +isigma$ to $beta$.
– Paul Enta
Jul 19 at 9:50
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
Okay, that helps. Thanks!
– Florian
Jul 19 at 10:32
add a comment |Â
up vote
0
down vote
Consider function:
$$y=sqrtx^2+a^2 ln e^sqrt x^2 +a^2$$
$$y'=fracxsqrt x^2+a^2.(e^ sqrtx^2+a^2)(1+sqrtx^2+a^2)$$
Now compare this with the relation in question we get:
$$k=-[frac ln x(1+sqrtx^2+a^2)sqrt x^2+a^2+1]i$$
This is condition for initial relation to be integrable.
answered Jul 18 at 17:35
sirous
761511
add a comment |Â
up vote
0
down vote
Consider function:
$$y=sqrtx^2+a^2 ln e^sqrt x^2 +a^2$$
$$y'=fracxsqrt x^2+a^2.(e^ sqrtx^2+a^2)(1+sqrtx^2+a^2)$$
Now compare this with the relation in question we get:
$$k=-[frac ln x(1+sqrtx^2+a^2)sqrt x^2+a^2+1]i$$
This is condition for initial relation to be integrable.
answered Jul 18 at 17:35
sirous
761511
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider function:
$$y=sqrtx^2+a^2 ln e^sqrt x^2 +a^2$$
$$y'=fracxsqrt x^2+a^2.(e^ sqrtx^2+a^2)(1+sqrtx^2+a^2)$$
Now compare this with the relation in question we get:
$$k=-[frac ln x(1+sqrtx^2+a^2)sqrt x^2+a^2+1]i$$
This is condition for initial relation to be integrable.
answered Jul 18 at 17:35
sirous
761511
Consider function:
$$y=sqrtx^2+a^2 ln e^sqrt x^2 +a^2$$
$$y'=fracxsqrt x^2+a^2.(e^ sqrtx^2+a^2)(1+sqrtx^2+a^2)$$
Now compare this with the relation in question we get:
$$k=-[frac ln x(1+sqrtx^2+a^2)sqrt x^2+a^2+1]i$$
This is condition for initial relation to be integrable.
answered Jul 18 at 17:35
sirous
761511
edited Jul 18 at 18:00
answered Jul 18 at 17:35
sirous
761511
answered Jul 18 at 17:35
sirous
761511
answered Jul 18 at 17:35
sirous
761511
761511
add a comment |Â
add a comment |Â
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Looks like something Bessel-ish.
– Frpzzd
Jul 18 at 11:48
Yeah, I was thinking the same. The Exp with something trigonometric-ish inside calls for Bessel. But I couldn't bring it in any Bessel integral form.
– Florian
Jul 18 at 11:50
If we have a complex component then an integral should have a defined integration path on the Argand plain. Is one defined?
– Dr Peter McGowan
Jul 18 at 11:50
I'm integrating over the real variable $x$. The integrand is complex but the integration variable is not. Might as well integrate real and imaginary part separately, in this case the question would be about $int fraccos[ksqrtx^2+a^2]sqrtx^2+a^2 rm dx$. This would help as well.
– Florian
Jul 18 at 11:52
With limits $-infty$ and $infty$ it is a standard integral, giving a Hankel function: dlmf.nist.gov/10.9.E10
– Chappers
Jul 18 at 12:22