Conditions for integrations by parts
$begingroup$
Let's say I have two periodic continuous functions $f(x)$,$g(x)$ with period $2pi $, which are also piecewise differentiable on the real line.
Does the following formula for integration by parts take hold? $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
I believe it does.
Here's my attempt of a proof:
Let's say ${x_1},...,{x_n} in left( { - pi ,pi } right)$ are the points in which either $f(x)$ or $g(x)$ is not differentiable (but the one sided derivatives exist). We define ${x_0} = - pi $ and ${x_{n + 1}} = pi $.
Then we can say (from the linearity of the integral) that $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } $$
and because $f(x)$ and $g(x)$ are continuously differentiable on each segment $left( {{x_{k - 1}},{x_k}} right)$ and are continuous on $left[ { - pi ,pi } right]$ we can use the regular integration by parts formula:
$$eqalign{
& intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } cr
& {rm{ }} = sumlimits_{k = 1}^{n + 1} {left[ {fleft( {{x_k}} right)gleft( {{x_k}} right) - fleft( {{x_{k - 1}}} right)gleft( {{x_{k - 1}}} right) - intlimits_{{x_{k - 1}}}^{{x_k}} {f(x)g'(x)dx} } right]} cr} $$
There are a lot of terms which cancel out and we are left with
$$left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
QED.
Is my proof correct?
Thank you a lot in advance!
calculus proof-verification
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
$endgroup$
add a comment |
$begingroup$
Let's say I have two periodic continuous functions $f(x)$,$g(x)$ with period $2pi $, which are also piecewise differentiable on the real line.
Does the following formula for integration by parts take hold? $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
I believe it does.
Here's my attempt of a proof:
Let's say ${x_1},...,{x_n} in left( { - pi ,pi } right)$ are the points in which either $f(x)$ or $g(x)$ is not differentiable (but the one sided derivatives exist). We define ${x_0} = - pi $ and ${x_{n + 1}} = pi $.
Then we can say (from the linearity of the integral) that $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } $$
and because $f(x)$ and $g(x)$ are continuously differentiable on each segment $left( {{x_{k - 1}},{x_k}} right)$ and are continuous on $left[ { - pi ,pi } right]$ we can use the regular integration by parts formula:
$$eqalign{
& intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } cr
& {rm{ }} = sumlimits_{k = 1}^{n + 1} {left[ {fleft( {{x_k}} right)gleft( {{x_k}} right) - fleft( {{x_{k - 1}}} right)gleft( {{x_{k - 1}}} right) - intlimits_{{x_{k - 1}}}^{{x_k}} {f(x)g'(x)dx} } right]} cr} $$
There are a lot of terms which cancel out and we are left with
$$left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
QED.
Is my proof correct?
Thank you a lot in advance!
calculus proof-verification
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
$endgroup$
add a comment |
$begingroup$
Let's say I have two periodic continuous functions $f(x)$,$g(x)$ with period $2pi $, which are also piecewise differentiable on the real line.
Does the following formula for integration by parts take hold? $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
I believe it does.
Here's my attempt of a proof:
Let's say ${x_1},...,{x_n} in left( { - pi ,pi } right)$ are the points in which either $f(x)$ or $g(x)$ is not differentiable (but the one sided derivatives exist). We define ${x_0} = - pi $ and ${x_{n + 1}} = pi $.
Then we can say (from the linearity of the integral) that $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } $$
and because $f(x)$ and $g(x)$ are continuously differentiable on each segment $left( {{x_{k - 1}},{x_k}} right)$ and are continuous on $left[ { - pi ,pi } right]$ we can use the regular integration by parts formula:
$$eqalign{
& intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } cr
& {rm{ }} = sumlimits_{k = 1}^{n + 1} {left[ {fleft( {{x_k}} right)gleft( {{x_k}} right) - fleft( {{x_{k - 1}}} right)gleft( {{x_{k - 1}}} right) - intlimits_{{x_{k - 1}}}^{{x_k}} {f(x)g'(x)dx} } right]} cr} $$
There are a lot of terms which cancel out and we are left with
$$left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
QED.
Is my proof correct?
Thank you a lot in advance!
calculus proof-verification
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
$endgroup$
Let's say I have two periodic continuous functions $f(x)$,$g(x)$ with period $2pi $, which are also piecewise differentiable on the real line.
Does the following formula for integration by parts take hold? $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
I believe it does.
Here's my attempt of a proof:
Let's say ${x_1},...,{x_n} in left( { - pi ,pi } right)$ are the points in which either $f(x)$ or $g(x)$ is not differentiable (but the one sided derivatives exist). We define ${x_0} = - pi $ and ${x_{n + 1}} = pi $.
Then we can say (from the linearity of the integral) that $$intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } $$
and because $f(x)$ and $g(x)$ are continuously differentiable on each segment $left( {{x_{k - 1}},{x_k}} right)$ and are continuous on $left[ { - pi ,pi } right]$ we can use the regular integration by parts formula:
$$eqalign{
& intlimits_{ - pi }^pi {f'(x)g(x)dx} = sumlimits_{k = 1}^{n + 1} {intlimits_{{x_{k - 1}}}^{{x_k}} {f'(x)g(x)dx} } cr
& {rm{ }} = sumlimits_{k = 1}^{n + 1} {left[ {fleft( {{x_k}} right)gleft( {{x_k}} right) - fleft( {{x_{k - 1}}} right)gleft( {{x_{k - 1}}} right) - intlimits_{{x_{k - 1}}}^{{x_k}} {f(x)g'(x)dx} } right]} cr} $$
There are a lot of terms which cancel out and we are left with
$$left. {f(x)g(x)} right|_{ - pi }^pi - intlimits_{ - pi }^pi {f(x)g'(x)dx} $$
QED.
Is my proof correct?
Thank you a lot in advance!
calculus proof-verification
calculus proof-verification
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
asked Dec 5 '18 at 14:57
zokomokozokomoko
162214
162214
add a comment |
add a comment |
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