calculate exponential integral
0
$begingroup$
$$int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx$$
and $0<a_1<b_1,d>0，c>0$. How to calculate this integral? Or a lower bound of this integral?
integration
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
$endgroup$
add a comment |
0
$begingroup$
$$int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx$$
and $0<a_1<b_1,d>0，c>0$. How to calculate this integral? Or a lower bound of this integral?
integration
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
$endgroup$
$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42
add a comment |
0
0
0
$begingroup$
$$int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx$$
and $0<a_1<b_1,d>0，c>0$. How to calculate this integral? Or a lower bound of this integral?
integration
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
$endgroup$
$$int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx$$
and $0<a_1<b_1,d>0，c>0$. How to calculate this integral? Or a lower bound of this integral?
integration
integration
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
edited Dec 23 '18 at 4:19
gt6989b
35.2k22557
35.2k22557
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
asked Dec 23 '18 at 3:51
Victor.XiaoVictor.Xiao
1
1
$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42
add a comment |
$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42
$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42
add a comment |
0
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$begingroup$
So you have $$ begin{split} I &= int_0^cleft(e^{-ax}- e^{-bx}right)e^{-d/x}dx \ &= int_0^c e^{-ax - d/x}dx - int_0^c e^{-bx - d/x}dx \ &= f(a,c,d) - f(b,c,d), end{split} $$ which reduces the problem to finding $$ f(a,c,d) = int_0^c e^{-ax - d/x}dx $$
$endgroup$
– gt6989b
Dec 23 '18 at 4:23
$begingroup$
Yes, you are right. But I failed in solving this sub-problem. It is integrable but the exact expression may be hard to derive. Hence, I am looking for a lower bound of the original problem.
$endgroup$
– Victor.Xiao
Dec 23 '18 at 4:42