Mixing
Maximizing a certain integral
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let $f:[0,1]to [0,infty)$ and $g:[0,1]to (0,infty)$ be smooth functions ($g$ is strictly positive). Define $varphi:(0,infty)to (0,infty)$ by $$varphi(t)=t^-frac12int_0^1 g(x)e^-fracf(x)tdx.$$
Suppose we know that $varphi$ attains a maximum (for example, when $f$ and $g$ are constant), and let it be attained at $t^astin (0,infty)$, i.e. $$varphi(t^ast)=max_tin (0,infty)varphi(t).$$
What is the maximum of $varphi$ and what is $t^ast$ in terms of $f$ and $g$?
My attempt: differentiate under the integral sign: $$fracdvarphidt=frac12t^5/2int_0^1(2f(x)-t)g(x)e^-fracf(x)tdx.$$ Therefore $t$ is a maximum iff $dvarphi/dt=0$ which is true iff $$t=frac2int_0^1f(x)g(x)e^-fracf(x)tdxint_0^1g(x)e^-fracf(x)tdx.$$ Now I'm stuck. How can I solve for t?
Wolfram spat out something about the Lambert $W$ function, but I'm not familiar with that. Thanks for your help.
real-analysis integration
edited Jul 16 at 12:06
md2perpe
5,95511022
asked Jul 16 at 2:16
smnas
23017
add a comment |Â
up vote
0
down vote
favorite
Let $f:[0,1]to [0,infty)$ and $g:[0,1]to (0,infty)$ be smooth functions ($g$ is strictly positive). Define $varphi:(0,infty)to (0,infty)$ by $$varphi(t)=t^-frac12int_0^1 g(x)e^-fracf(x)tdx.$$
Suppose we know that $varphi$ attains a maximum (for example, when $f$ and $g$ are constant), and let it be attained at $t^astin (0,infty)$, i.e. $$varphi(t^ast)=max_tin (0,infty)varphi(t).$$
What is the maximum of $varphi$ and what is $t^ast$ in terms of $f$ and $g$?
My attempt: differentiate under the integral sign: $$fracdvarphidt=frac12t^5/2int_0^1(2f(x)-t)g(x)e^-fracf(x)tdx.$$ Therefore $t$ is a maximum iff $dvarphi/dt=0$ which is true iff $$t=frac2int_0^1f(x)g(x)e^-fracf(x)tdxint_0^1g(x)e^-fracf(x)tdx.$$ Now I'm stuck. How can I solve for t?
Wolfram spat out something about the Lambert $W$ function, but I'm not familiar with that. Thanks for your help.
real-analysis integration
edited Jul 16 at 12:06
md2perpe
5,95511022
asked Jul 16 at 2:16
smnas
23017
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
I made it up myself
– smnas
Jul 19 at 14:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:[0,1]to [0,infty)$ and $g:[0,1]to (0,infty)$ be smooth functions ($g$ is strictly positive). Define $varphi:(0,infty)to (0,infty)$ by $$varphi(t)=t^-frac12int_0^1 g(x)e^-fracf(x)tdx.$$
Suppose we know that $varphi$ attains a maximum (for example, when $f$ and $g$ are constant), and let it be attained at $t^astin (0,infty)$, i.e. $$varphi(t^ast)=max_tin (0,infty)varphi(t).$$
What is the maximum of $varphi$ and what is $t^ast$ in terms of $f$ and $g$?
My attempt: differentiate under the integral sign: $$fracdvarphidt=frac12t^5/2int_0^1(2f(x)-t)g(x)e^-fracf(x)tdx.$$ Therefore $t$ is a maximum iff $dvarphi/dt=0$ which is true iff $$t=frac2int_0^1f(x)g(x)e^-fracf(x)tdxint_0^1g(x)e^-fracf(x)tdx.$$ Now I'm stuck. How can I solve for t?
Wolfram spat out something about the Lambert $W$ function, but I'm not familiar with that. Thanks for your help.
real-analysis integration
edited Jul 16 at 12:06
md2perpe
5,95511022
asked Jul 16 at 2:16
smnas
23017
Let $f:[0,1]to [0,infty)$ and $g:[0,1]to (0,infty)$ be smooth functions ($g$ is strictly positive). Define $varphi:(0,infty)to (0,infty)$ by $$varphi(t)=t^-frac12int_0^1 g(x)e^-fracf(x)tdx.$$
Suppose we know that $varphi$ attains a maximum (for example, when $f$ and $g$ are constant), and let it be attained at $t^astin (0,infty)$, i.e. $$varphi(t^ast)=max_tin (0,infty)varphi(t).$$
What is the maximum of $varphi$ and what is $t^ast$ in terms of $f$ and $g$?
My attempt: differentiate under the integral sign: $$fracdvarphidt=frac12t^5/2int_0^1(2f(x)-t)g(x)e^-fracf(x)tdx.$$ Therefore $t$ is a maximum iff $dvarphi/dt=0$ which is true iff $$t=frac2int_0^1f(x)g(x)e^-fracf(x)tdxint_0^1g(x)e^-fracf(x)tdx.$$ Now I'm stuck. How can I solve for t?
Wolfram spat out something about the Lambert $W$ function, but I'm not familiar with that. Thanks for your help.
real-analysis integration
edited Jul 16 at 12:06
md2perpe
5,95511022
asked Jul 16 at 2:16
smnas
23017
edited Jul 16 at 12:06
md2perpe
5,95511022
edited Jul 16 at 12:06
md2perpe
5,95511022
edited Jul 16 at 12:06
md2perpe
5,95511022
5,95511022
asked Jul 16 at 2:16
smnas
23017
asked Jul 16 at 2:16
smnas
23017
asked Jul 16 at 2:16
smnas
23017
23017
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
I made it up myself
– smnas
Jul 19 at 14:09
add a comment |Â
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
I made it up myself
– smnas
Jul 19 at 14:09
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
I made it up myself
– smnas
Jul 19 at 14:09
I made it up myself
– smnas
Jul 19 at 14:09
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853009%2fmaximizing-a-certain-integral%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Where did you find this problem?
– md2perpe
Jul 16 at 13:05
I made it up myself
– smnas
Jul 19 at 14:09