If $a$ and $b$ are constants, calculate the definite integral
$begingroup$
$$int_{-infty}^{+infty} F(x-b) f(x-a) dx$$
$$f(x) = exp(-x-e^{-x}), qquad x in (-infty, +infty)$$
$$F(x) = int_{-infty}^x f(t) dt$$
I calculated already the integral of $F(x)$, which is $exp(-e^{-x})$, but I am stuck on the other one, I have no idea how to calculate the integral of $F(x-a)f(x-b)$.
integration
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
$endgroup$
add a comment |
$begingroup$
$$int_{-infty}^{+infty} F(x-b) f(x-a) dx$$
$$f(x) = exp(-x-e^{-x}), qquad x in (-infty, +infty)$$
$$F(x) = int_{-infty}^x f(t) dt$$
I calculated already the integral of $F(x)$, which is $exp(-e^{-x})$, but I am stuck on the other one, I have no idea how to calculate the integral of $F(x-a)f(x-b)$.
integration
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26
add a comment |
$begingroup$
$$int_{-infty}^{+infty} F(x-b) f(x-a) dx$$
$$f(x) = exp(-x-e^{-x}), qquad x in (-infty, +infty)$$
$$F(x) = int_{-infty}^x f(t) dt$$
I calculated already the integral of $F(x)$, which is $exp(-e^{-x})$, but I am stuck on the other one, I have no idea how to calculate the integral of $F(x-a)f(x-b)$.
integration
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
$endgroup$
$$int_{-infty}^{+infty} F(x-b) f(x-a) dx$$
$$f(x) = exp(-x-e^{-x}), qquad x in (-infty, +infty)$$
$$F(x) = int_{-infty}^x f(t) dt$$
I calculated already the integral of $F(x)$, which is $exp(-e^{-x})$, but I am stuck on the other one, I have no idea how to calculate the integral of $F(x-a)f(x-b)$.
integration
integration
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
edited Dec 28 '18 at 8:39
Rócherz
3,0263823
3,0263823
asked Dec 28 '18 at 8:22
user10471408user10471408
162
asked Dec 28 '18 at 8:22
user10471408user10471408
162
asked Dec 28 '18 at 8:22
user10471408user10471408
162
162
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26
add a comment |
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26
add a comment |
1 Answer
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$begingroup$
$$int_{-infty}^{infty}F(a-b)f(x-a)dx=int_{-infty}^{infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$
$Rightarrow e^{a} int_{-infty}^{infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$
Now Let, $-e^{-x}=t, rightarrow e^{-x} dx =dt$
$Rightarrow e^{a} int_{-infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$
$Rightarrow frac{e^a}{e^{a}+e^{b}}.$
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$int_{-infty}^{infty}F(a-b)f(x-a)dx=int_{-infty}^{infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$
$Rightarrow e^{a} int_{-infty}^{infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$
Now Let, $-e^{-x}=t, rightarrow e^{-x} dx =dt$
$Rightarrow e^{a} int_{-infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$
$Rightarrow frac{e^a}{e^{a}+e^{b}}.$
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
$endgroup$
add a comment |
$begingroup$
$$int_{-infty}^{infty}F(a-b)f(x-a)dx=int_{-infty}^{infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$
$Rightarrow e^{a} int_{-infty}^{infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$
Now Let, $-e^{-x}=t, rightarrow e^{-x} dx =dt$
$Rightarrow e^{a} int_{-infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$
$Rightarrow frac{e^a}{e^{a}+e^{b}}.$
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
$endgroup$
add a comment |
$begingroup$
$$int_{-infty}^{infty}F(a-b)f(x-a)dx=int_{-infty}^{infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$
$Rightarrow e^{a} int_{-infty}^{infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$
Now Let, $-e^{-x}=t, rightarrow e^{-x} dx =dt$
$Rightarrow e^{a} int_{-infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$
$Rightarrow frac{e^a}{e^{a}+e^{b}}.$
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
$endgroup$
$$int_{-infty}^{infty}F(a-b)f(x-a)dx=int_{-infty}^{infty}e^{-e^{-(x-b)}} e^{-(x-a)} e^{-e^{-(x-a)}} dx=$$
$Rightarrow e^{a} int_{-infty}^{infty} e^{-e^{-x}({e^{b}+e^{a}})} e^{-x} dx$
Now Let, $-e^{-x}=t, rightarrow e^{-x} dx =dt$
$Rightarrow e^{a} int_{-infty}^{0} e^{t{(e^{a}+e^{b}})} dt~~$
$Rightarrow frac{e^a}{e^{a}+e^{b}}.$
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
edited Dec 28 '18 at 19:02
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
answered Dec 28 '18 at 10:05
Sachin KumarSachin Kumar
245110
245110
add a comment |
add a comment |
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$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Dec 28 '18 at 8:26