Mixing
Integrals with the special functions $Ci(x)$ and $erf(x)$
$begingroup$
I'm looking for the solutions of the following two integrals:
$$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
and
$$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
with
$$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
and
$$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
and
$$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?
calculus real-analysis integration definite-integrals special-functions
edited Jan 15 at 2:30
Henry Lee
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
$endgroup$
add a comment |
$begingroup$
I'm looking for the solutions of the following two integrals:
$$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
and
$$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
with
$$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
and
$$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
and
$$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?
calculus real-analysis integration definite-integrals special-functions
edited Jan 15 at 2:30
Henry Lee
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
$endgroup$
add a comment |
$begingroup$
I'm looking for the solutions of the following two integrals:
$$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
and
$$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
with
$$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
and
$$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
and
$$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?
calculus real-analysis integration definite-integrals special-functions
edited Jan 15 at 2:30
Henry Lee
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
$endgroup$
I'm looking for the solutions of the following two integrals:
$$I_1=intlimits_0^infty dx, e^{-x^2}text{Ci}(ax)$$
and
$$I_2=intlimits_0^infty dx, e^{-ax}text{erf}(x)$$
with
$$text{Ci}(x)=intlimits_infty^xfrac{cos(t)}{t}dt$$
and
$$text{erf}(x)=frac{2}{sqrt{pi}}intlimits_0^xe^{-t^2}dt$$
Now I'm not 100% sure what is meant by "the solution of the integrals" since these will probably be not-evaluable. But I'm guessing that the question is to reduce the expression to one single special function in stead of the integral of a special function with an elementary function.
Mathematica yields me the answers:
$$I_1=-frac{sqrt{pi}}{4}Gammaleft(0,frac{a^2}{4}right)$$
and
$$I_2=expleft(frac{a^2}{4}right)frac{1-text{erf}(a/2)}{a}$$
A good first step for evaluating these integrals $I_1$ and $I_2$ seemed to fill in the integral representations of these special functions and try to switch te integrals over $x$ and $t$. However this has not (yet) been a success. I also tried to find a differential equation for these integrals, but also this was not so easy to do. Are there any tips/tricks to evaluate these integrals ?
calculus real-analysis integration definite-integrals special-functions
calculus real-analysis integration definite-integrals special-functions
edited Jan 15 at 2:30
Henry Lee
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
edited Jan 15 at 2:30
Henry Lee
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
edited Jan 15 at 2:30
Henry Lee
2,042219
edited Jan 15 at 2:30
Henry Lee
2,042219
edited Jan 15 at 2:30
Henry Lee
2,042219
2,042219
asked Mar 23 '15 at 12:20
NickNick
494518
asked Mar 23 '15 at 12:20
NickNick
494518
asked Mar 23 '15 at 12:20
NickNick
494518
494518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
$endgroup$
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
add a comment |
$begingroup$
For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=frac{a}{2}+x$
This turns into
$$
I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
$$
As promised!
I will have a look on $I_1$ later
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
$endgroup$
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
add a comment |
$begingroup$
Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
$endgroup$
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
add a comment |
$begingroup$
Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
$endgroup$
Hint: Differentiate $I_1(a)$ with regard to a. Similarly, define $~I_2(b)~=~displaystyleint_0^infty e^{-ax}~text{erf}(bx)~dx,~$ and then differentiate it with regard to b.
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
answered Mar 23 '15 at 13:27
LucianLucian
41.2k159130
41.2k159130
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
add a comment |
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
$begingroup$
Nice, adding a factor in order to eliminate one of the integrations !
$endgroup$
– Nick
Mar 23 '15 at 14:51
add a comment |
$begingroup$
For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=frac{a}{2}+x$
This turns into
$$
I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
$$
As promised!
I will have a look on $I_1$ later
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
add a comment |
$begingroup$
For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=frac{a}{2}+x$
This turns into
$$
I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
$$
As promised!
I will have a look on $I_1$ later
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
$endgroup$
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
add a comment |
$begingroup$
For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=frac{a}{2}+x$
This turns into
$$
I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
$$
As promised!
I will have a look on $I_1$ later
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
$endgroup$
For $I_2$, please note that this integral is just the Laplace transform of $text{Erf}(x)$ w.r.t. to the variable $a$
Furthermore the Laplace transform of an integral $mathcal{L}(int f(x) dx)$ is just $mathcal{L}(f(x))/a$
Using this we find that
$$
I_2=frac{2}{asqrt{pi}}int_0^{infty}e^{- a x}e^{-x^2}dx
$$
Completing the square in the exponent and changing variables $y=frac{a}{2}+x$
This turns into
$$
I_2=2frac{ e^{frac{a^2}{4}}}{asqrt{pi}}int_{frac{a}{2}}^{infty}e^{-y^2}dy
$$
Now using straightforwardly the definition of the Errorfunction we obtain
$$
I_2=frac{e^{frac{a^2}{4}}}{a}left(1-text{Erf}(a/2)right)
$$
As promised!
I will have a look on $I_1$ later
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
answered Mar 23 '15 at 13:24
tiredtired
10.6k12043
10.6k12043
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
add a comment |
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
$begingroup$
thanks a lot, that's already a nice first step. I somehow have the feeling that both integrals $I_1$ and $I_2$ should be analogous (due to a slight analogy in the answer). So I didn't consider Laplace because it does not appear in the first integral.
$endgroup$
– Nick
Mar 23 '15 at 13:27
add a comment |
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