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Quick question why setting $a=0$ gives an indeterminate solution
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$newcommanddxmathrm dx,$I’m having lots of trouble figuring this out, so perhaps you guys can help me. For example, let’s take the integral
$$intlimits_0^inftydxfrac sin xlog xx=-frac gammapi2$$
Our integral can be computed using differentiation under the integral sign and taking the imaginary part of
$$mathfrakI(s)=-intlimits_0^inftydx x^a-1e^-sx=-s^-aGamma(a)$$
Set $s=i$ and take the imaginary part to get
$$operatornameImmathfrakI(i)=Gamma(a)sinfrac pi a2$$
But when I differentiate and set $a=0$, then the gamma function becomes undefined because $Gamma’(0)$ doesn’t produce a determinate form.
I’m not exactly sure what went wrong. Perhaps you can help me?
integration
asked Jul 30 at 3:04
Crescendo
2,2691525
add a comment |Â
up vote
3
down vote
favorite
$newcommanddxmathrm dx,$I’m having lots of trouble figuring this out, so perhaps you guys can help me. For example, let’s take the integral
$$intlimits_0^inftydxfrac sin xlog xx=-frac gammapi2$$
Our integral can be computed using differentiation under the integral sign and taking the imaginary part of
$$mathfrakI(s)=-intlimits_0^inftydx x^a-1e^-sx=-s^-aGamma(a)$$
Set $s=i$ and take the imaginary part to get
$$operatornameImmathfrakI(i)=Gamma(a)sinfrac pi a2$$
But when I differentiate and set $a=0$, then the gamma function becomes undefined because $Gamma’(0)$ doesn’t produce a determinate form.
I’m not exactly sure what went wrong. Perhaps you can help me?
integration
asked Jul 30 at 3:04
Crescendo
2,2691525
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$newcommanddxmathrm dx,$I’m having lots of trouble figuring this out, so perhaps you guys can help me. For example, let’s take the integral
$$intlimits_0^inftydxfrac sin xlog xx=-frac gammapi2$$
Our integral can be computed using differentiation under the integral sign and taking the imaginary part of
$$mathfrakI(s)=-intlimits_0^inftydx x^a-1e^-sx=-s^-aGamma(a)$$
Set $s=i$ and take the imaginary part to get
$$operatornameImmathfrakI(i)=Gamma(a)sinfrac pi a2$$
But when I differentiate and set $a=0$, then the gamma function becomes undefined because $Gamma’(0)$ doesn’t produce a determinate form.
I’m not exactly sure what went wrong. Perhaps you can help me?
integration
asked Jul 30 at 3:04
Crescendo
2,2691525
$newcommanddxmathrm dx,$I’m having lots of trouble figuring this out, so perhaps you guys can help me. For example, let’s take the integral
$$intlimits_0^inftydxfrac sin xlog xx=-frac gammapi2$$
Our integral can be computed using differentiation under the integral sign and taking the imaginary part of
$$mathfrakI(s)=-intlimits_0^inftydx x^a-1e^-sx=-s^-aGamma(a)$$
Set $s=i$ and take the imaginary part to get
$$operatornameImmathfrakI(i)=Gamma(a)sinfrac pi a2$$
But when I differentiate and set $a=0$, then the gamma function becomes undefined because $Gamma’(0)$ doesn’t produce a determinate form.
I’m not exactly sure what went wrong. Perhaps you can help me?
integration
asked Jul 30 at 3:04
Crescendo
2,2691525
asked Jul 30 at 3:04
Crescendo
2,2691525
asked Jul 30 at 3:04
Crescendo
2,2691525
asked Jul 30 at 3:04
Crescendo
2,2691525
2,2691525
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
From Laplace transform we know that
$$int_0^infty e^-sxx^a-1dx=dfracGamma(a)s^a$$
is valid where $bf Re(s)>0$, so your reasoning about substitution $s=i$ is false here.
answered Jul 30 at 3:41
user 108128
19k41544
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
From Laplace transform we know that
$$int_0^infty e^-sxx^a-1dx=dfracGamma(a)s^a$$
is valid where $bf Re(s)>0$, so your reasoning about substitution $s=i$ is false here.
answered Jul 30 at 3:41
user 108128
19k41544
add a comment |Â
up vote
0
down vote
From Laplace transform we know that
$$int_0^infty e^-sxx^a-1dx=dfracGamma(a)s^a$$
is valid where $bf Re(s)>0$, so your reasoning about substitution $s=i$ is false here.
answered Jul 30 at 3:41
user 108128
19k41544
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From Laplace transform we know that
$$int_0^infty e^-sxx^a-1dx=dfracGamma(a)s^a$$
is valid where $bf Re(s)>0$, so your reasoning about substitution $s=i$ is false here.
answered Jul 30 at 3:41
user 108128
19k41544
From Laplace transform we know that
$$int_0^infty e^-sxx^a-1dx=dfracGamma(a)s^a$$
is valid where $bf Re(s)>0$, so your reasoning about substitution $s=i$ is false here.
answered Jul 30 at 3:41
user 108128
19k41544
answered Jul 30 at 3:41
user 108128
19k41544
answered Jul 30 at 3:41
user 108128
19k41544
answered Jul 30 at 3:41
user 108128
19k41544
19k41544
add a comment |Â
add a comment |Â
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