Is this way for integrating has any fault?
$begingroup$
I have tried this $int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx $
integration in the method below-
$$int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx\=int_{-infty}^{0}e^{-2(-x)/a}cdot frac{d^2}{dx^2}(e^{-2(-x)/a})~~dx +int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=int_{-infty}^{0}e^{2x/a}cdot frac{d^2}{dx^2}(e^{2x/a})~~dx+int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=4int_{-infty}^{0}e^{4x/a}~~dx+4int_{0}^{infty}e^{-4x/a}~~dx\
=4cdot frac{a}{4}+4cdot frac a4=2a$$
I am not sure I am correct or not. Please help me with this integral and correct me if I am wrong at any step. Any help will be appreciated.
integration definite-integrals
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
$endgroup$
add a comment |
$begingroup$
I have tried this $int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx $
integration in the method below-
$$int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx\=int_{-infty}^{0}e^{-2(-x)/a}cdot frac{d^2}{dx^2}(e^{-2(-x)/a})~~dx +int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=int_{-infty}^{0}e^{2x/a}cdot frac{d^2}{dx^2}(e^{2x/a})~~dx+int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=4int_{-infty}^{0}e^{4x/a}~~dx+4int_{0}^{infty}e^{-4x/a}~~dx\
=4cdot frac{a}{4}+4cdot frac a4=2a$$
I am not sure I am correct or not. Please help me with this integral and correct me if I am wrong at any step. Any help will be appreciated.
integration definite-integrals
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
$endgroup$
1
$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17
add a comment |
$begingroup$
I have tried this $int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx $
integration in the method below-
$$int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx\=int_{-infty}^{0}e^{-2(-x)/a}cdot frac{d^2}{dx^2}(e^{-2(-x)/a})~~dx +int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=int_{-infty}^{0}e^{2x/a}cdot frac{d^2}{dx^2}(e^{2x/a})~~dx+int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=4int_{-infty}^{0}e^{4x/a}~~dx+4int_{0}^{infty}e^{-4x/a}~~dx\
=4cdot frac{a}{4}+4cdot frac a4=2a$$
I am not sure I am correct or not. Please help me with this integral and correct me if I am wrong at any step. Any help will be appreciated.
integration definite-integrals
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
$endgroup$
I have tried this $int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx $
integration in the method below-
$$int_{-infty}^{infty}e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})~~dx\=int_{-infty}^{0}e^{-2(-x)/a}cdot frac{d^2}{dx^2}(e^{-2(-x)/a})~~dx +int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=int_{-infty}^{0}e^{2x/a}cdot frac{d^2}{dx^2}(e^{2x/a})~~dx+int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a})~~dx\
=4int_{-infty}^{0}e^{4x/a}~~dx+4int_{0}^{infty}e^{-4x/a}~~dx\
=4cdot frac{a}{4}+4cdot frac a4=2a$$
I am not sure I am correct or not. Please help me with this integral and correct me if I am wrong at any step. Any help will be appreciated.
integration definite-integrals
integration definite-integrals
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
asked Jan 11 at 5:07
OppoInfinityOppoInfinity
18911
18911
1
$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17
add a comment |
1
$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17
1
1
$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17
$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17
add a comment |
1 Answer
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$begingroup$
Let $a>0$. Function $f=e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})$ is even. Then
$$int_{-infty}^{infty}f,dx=2int_{0}^{infty}f,dx\
=2int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a}),dx\
=2int_{0}^{infty}frac{4 {e^{-frac{4 x}{a}}}}{{{a}^{2}}}dx=frac2a$$
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
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add a comment |
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$begingroup$
Let $a>0$. Function $f=e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})$ is even. Then
$$int_{-infty}^{infty}f,dx=2int_{0}^{infty}f,dx\
=2int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a}),dx\
=2int_{0}^{infty}frac{4 {e^{-frac{4 x}{a}}}}{{{a}^{2}}}dx=frac2a$$
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let $a>0$. Function $f=e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})$ is even. Then
$$int_{-infty}^{infty}f,dx=2int_{0}^{infty}f,dx\
=2int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a}),dx\
=2int_{0}^{infty}frac{4 {e^{-frac{4 x}{a}}}}{{{a}^{2}}}dx=frac2a$$
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let $a>0$. Function $f=e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})$ is even. Then
$$int_{-infty}^{infty}f,dx=2int_{0}^{infty}f,dx\
=2int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a}),dx\
=2int_{0}^{infty}frac{4 {e^{-frac{4 x}{a}}}}{{{a}^{2}}}dx=frac2a$$
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
Let $a>0$. Function $f=e^{-2|x|/a}cdot frac{d^2}{dx^2}(e^{-2|x|/a})$ is even. Then
$$int_{-infty}^{infty}f,dx=2int_{0}^{infty}f,dx\
=2int_{0}^{infty}e^{-2x/a}cdot frac{d^2}{dx^2}(e^{-2x/a}),dx\
=2int_{0}^{infty}frac{4 {e^{-frac{4 x}{a}}}}{{{a}^{2}}}dx=frac2a$$
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 11 at 8:45
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
add a comment |
add a comment |
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$begingroup$
The derivatives bring down $4/a^{2}$, not just 4.
$endgroup$
– Ininterrompue
Jan 11 at 5:17