An integral leads to complementary error functions
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I am reading a paper Albrecher, Constantinescu and Loisel "2011Explicit ruin formulas for models with dependence among risks" and getting stuck at one integral (Example 2.4):
$$int _{frac{lambda }{c}}^{infty }frac{lambda }{theta c}e^{-u theta } e^{frac{lambda c}{ u}} frac{alpha }{2sqrt{pi
theta ^3}}e^{-frac{alpha ^2}{4theta }}dtheta.$$
The $theta^{-frac{5}{2}}$ in the integrand kills my attempts. Three authors claim that it should be:
begin{equation}
frac{lambda }{alpha ^2 c}e^{-frac{alpha ^2c}{4lambda }}left{-frac{2 alpha }{sqrt{pi frac{lambda }{c}}}+e^{frac{left(alpha
c-2 lambda sqrt{u}right)^2}{4c lambda }}left(1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda }{c}}-frac{alpha }{2sqrt{frac{lambda
}{c}}}right]+e^{frac{left(alpha c+2 lambda sqrt{u}right)^2}{4c lambda }}left(-1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda
}{c}}+frac{alpha }{2sqrt{frac{lambda }{c}}}right]right},
end{equation}
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.
I cannot see how to do this. I start from the above and end with the following:
$$frac{lambda }{alpha c}left{-frac{2 }{sqrt{pi frac{lambda }{c}}}e^{-frac{alpha ^2c}{4lambda }}+frac{ e^{frac{u lambda }{ c }}}{sqrt{pi
}}int _{frac{lambda }{c}}^{infty }e^{-u theta -frac{alpha ^2}{4theta }}text{ }left(frac{1}{theta sqrt{theta }}+frac{2u}{sqrt{theta
}}right)dtheta right}.$$
I cannot see why this is the first equation too. Any helps? Thanks in advance.
integration complex-analysis special-functions error-function
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
$endgroup$
add a comment |
$begingroup$
I am reading a paper Albrecher, Constantinescu and Loisel "2011Explicit ruin formulas for models with dependence among risks" and getting stuck at one integral (Example 2.4):
$$int _{frac{lambda }{c}}^{infty }frac{lambda }{theta c}e^{-u theta } e^{frac{lambda c}{ u}} frac{alpha }{2sqrt{pi
theta ^3}}e^{-frac{alpha ^2}{4theta }}dtheta.$$
The $theta^{-frac{5}{2}}$ in the integrand kills my attempts. Three authors claim that it should be:
begin{equation}
frac{lambda }{alpha ^2 c}e^{-frac{alpha ^2c}{4lambda }}left{-frac{2 alpha }{sqrt{pi frac{lambda }{c}}}+e^{frac{left(alpha
c-2 lambda sqrt{u}right)^2}{4c lambda }}left(1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda }{c}}-frac{alpha }{2sqrt{frac{lambda
}{c}}}right]+e^{frac{left(alpha c+2 lambda sqrt{u}right)^2}{4c lambda }}left(-1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda
}{c}}+frac{alpha }{2sqrt{frac{lambda }{c}}}right]right},
end{equation}
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.
I cannot see how to do this. I start from the above and end with the following:
$$frac{lambda }{alpha c}left{-frac{2 }{sqrt{pi frac{lambda }{c}}}e^{-frac{alpha ^2c}{4lambda }}+frac{ e^{frac{u lambda }{ c }}}{sqrt{pi
}}int _{frac{lambda }{c}}^{infty }e^{-u theta -frac{alpha ^2}{4theta }}text{ }left(frac{1}{theta sqrt{theta }}+frac{2u}{sqrt{theta
}}right)dtheta right}.$$
I cannot see why this is the first equation too. Any helps? Thanks in advance.
integration complex-analysis special-functions error-function
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
$endgroup$
$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
1
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
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Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41
add a comment |
$begingroup$
I am reading a paper Albrecher, Constantinescu and Loisel "2011Explicit ruin formulas for models with dependence among risks" and getting stuck at one integral (Example 2.4):
$$int _{frac{lambda }{c}}^{infty }frac{lambda }{theta c}e^{-u theta } e^{frac{lambda c}{ u}} frac{alpha }{2sqrt{pi
theta ^3}}e^{-frac{alpha ^2}{4theta }}dtheta.$$
The $theta^{-frac{5}{2}}$ in the integrand kills my attempts. Three authors claim that it should be:
begin{equation}
frac{lambda }{alpha ^2 c}e^{-frac{alpha ^2c}{4lambda }}left{-frac{2 alpha }{sqrt{pi frac{lambda }{c}}}+e^{frac{left(alpha
c-2 lambda sqrt{u}right)^2}{4c lambda }}left(1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda }{c}}-frac{alpha }{2sqrt{frac{lambda
}{c}}}right]+e^{frac{left(alpha c+2 lambda sqrt{u}right)^2}{4c lambda }}left(-1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda
}{c}}+frac{alpha }{2sqrt{frac{lambda }{c}}}right]right},
end{equation}
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.
I cannot see how to do this. I start from the above and end with the following:
$$frac{lambda }{alpha c}left{-frac{2 }{sqrt{pi frac{lambda }{c}}}e^{-frac{alpha ^2c}{4lambda }}+frac{ e^{frac{u lambda }{ c }}}{sqrt{pi
}}int _{frac{lambda }{c}}^{infty }e^{-u theta -frac{alpha ^2}{4theta }}text{ }left(frac{1}{theta sqrt{theta }}+frac{2u}{sqrt{theta
}}right)dtheta right}.$$
I cannot see why this is the first equation too. Any helps? Thanks in advance.
integration complex-analysis special-functions error-function
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
$endgroup$
I am reading a paper Albrecher, Constantinescu and Loisel "2011Explicit ruin formulas for models with dependence among risks" and getting stuck at one integral (Example 2.4):
$$int _{frac{lambda }{c}}^{infty }frac{lambda }{theta c}e^{-u theta } e^{frac{lambda c}{ u}} frac{alpha }{2sqrt{pi
theta ^3}}e^{-frac{alpha ^2}{4theta }}dtheta.$$
The $theta^{-frac{5}{2}}$ in the integrand kills my attempts. Three authors claim that it should be:
begin{equation}
frac{lambda }{alpha ^2 c}e^{-frac{alpha ^2c}{4lambda }}left{-frac{2 alpha }{sqrt{pi frac{lambda }{c}}}+e^{frac{left(alpha
c-2 lambda sqrt{u}right)^2}{4c lambda }}left(1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda }{c}}-frac{alpha }{2sqrt{frac{lambda
}{c}}}right]+e^{frac{left(alpha c+2 lambda sqrt{u}right)^2}{4c lambda }}left(-1+alpha sqrt{u}right)text{erfc}left[sqrt{frac{u lambda
}{c}}+frac{alpha }{2sqrt{frac{lambda }{c}}}right]right},
end{equation}
where $operatorname{erfc}(t) = frac{2}{sqrt{pi}} int_t^infty e^{-x^2} dx$.
I cannot see how to do this. I start from the above and end with the following:
$$frac{lambda }{alpha c}left{-frac{2 }{sqrt{pi frac{lambda }{c}}}e^{-frac{alpha ^2c}{4lambda }}+frac{ e^{frac{u lambda }{ c }}}{sqrt{pi
}}int _{frac{lambda }{c}}^{infty }e^{-u theta -frac{alpha ^2}{4theta }}text{ }left(frac{1}{theta sqrt{theta }}+frac{2u}{sqrt{theta
}}right)dtheta right}.$$
I cannot see why this is the first equation too. Any helps? Thanks in advance.
integration complex-analysis special-functions error-function
integration complex-analysis special-functions error-function
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
edited Dec 20 '18 at 0:46
gouwangzhangdong
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
asked Dec 20 '18 at 0:40
gouwangzhangdonggouwangzhangdong
788
788
$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
1
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
$begingroup$
Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41
add a comment |
$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
1
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
$begingroup$
Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41
$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
1
1
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
$begingroup$
Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41
add a comment |
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$begingroup$
Maybe you would want to simplify your problem a bit because not many MSE users will be interested after seeing the messy and huge expressions.
$endgroup$
– Szeto
Dec 20 '18 at 12:23
1
$begingroup$
There is a typo, the formula with $operatorname{erfc}$ is the correct value for $$int_{lambda/c}^infty frac lambda {theta c} e^{-u theta} e^{lambda u/c} frac alpha {2 sqrt{pi theta^3}} e^{-alpha^2/(4 theta)} dtheta$$ ($e^{lambda u/c}$ instead of $e^{lambda c/u}$). For a given $k$, a closed form for $int theta^{k + 1/2} e^{-a theta - b/theta} dtheta$ exists because $$int e^{-(a x + b/x)^2} dx = frac {sqrt pi} {4 a} left( e^{-4 a b} operatorname{erfc} left( frac b x - a x right) - operatorname{erfc} left( frac b x + a x right) right).$$
$endgroup$
– Maxim
Dec 20 '18 at 14:45
$begingroup$
Yes，I just realise this typo. Many thanks for this.
$endgroup$
– gouwangzhangdong
Dec 21 '18 at 2:46
$begingroup$
This solution looks very similar, but I'm not sure if it helps: math.stackexchange.com/a/3051321/441161
$endgroup$
– Andy Walls
Dec 31 '18 at 3:41