# How to do Double integral with two function handle.

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gourav pandey 2021 年 8 月 13 日
コメント済み: Bjorn Gustavsson 2021 年 8 月 13 日
%%%%% theta and omega1 are two variables
temp1=@(theta,omega1)-omega1.^2.*exp(-abs(omega1).*(cos(theta) + 5)).*(((6632097892526661.*exp(-omega1))./2814749767106560 + (2351843784080109.*omega1.^(5./2).*((7055531352240327.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./(1125899906842624.*omega1.^(1./2)) + (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1).*(1./omega1 + 1))./4503599627370496))./9007199254740992 - (3527765676120163.*omega1.^(3./2).*((8466637622688393.*exp(-omega1).*(1./omega1 + 1))./(2251799813685248.*omega1.^(1./2)) + (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1))./4503599627370496))./4503599627370496 + 3.*pi.*exp(-omega1).*((5.*omega1)./2 + 5./2) + (696927317274246941282520301244943.*omega1.^2.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./50706024009129176059868128215040 - (31361729277341114213626645380177.*omega1.*exp(-omega1).*(1./omega1 + 1))./1267650600228229401496703205376 - (238946508779741850518491294873107.*omega1.^3.*exp(-omega1).*(6./omega1 + 15./omega1.^2 + 15./omega1.^3 + 1))./202824096036516704239472512860160 - (15.*pi.*omega1.*exp(-omega1))./2)./omega1.^2 - (cos(theta) + 5).*(((45860953789562119.*omega1.^(3./2).*((7055531352240327.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./(1125899906842624.*omega1.^(1./2)) - (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1).*(6./omega1 + 15./omega1.^2 + 15./omega1.^3 + 1))./4503599627370496))./11258999068426240 - (45860953789562119.*omega1.^(1./2).*((8466637622688393.*exp(-omega1).*(1./omega1 + 1))./(2251799813685248.*omega1.^(1./2)) - (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./4503599627370496))./5629499534213120 + (345144957126293784082265203705599.*omega1.^3.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./405648192073033408478945025720320 - (129429358922360136602258035107329.*omega1.^2.*exp(-omega1).*(1./omega1 + 1))./50706024009129176059868128215040)./omega1 - (abs(omega1).*((6632097892526661.*exp(-omega1))./2814749767106560 + (2351843784080109.*omega1.^(5./2).*((7055531352240327.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./(1125899906842624.*omega1.^(1./2)) + (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1).*(1./omega1 + 1))./4503599627370496))./9007199254740992 - (3527765676120163.*omega1.^(3./2).*((8466637622688393.*exp(-omega1).*(1./omega1 + 1))./(2251799813685248.*omega1.^(1./2)) + (3991211251234741.*2.^(1./2).*omega1.^(1./2).*exp(-omega1))./4503599627370496))./4503599627370496 + 3.*pi.*exp(-omega1).*((5.*omega1)./2 + 5./2) + (696927317274246941282520301244943.*omega1.^2.*exp(-omega1).*(3./omega1 + 3./omega1.^2 + 1))./50706024009129176059868128215040 - (31361729277341114213626645380177.*omega1.*exp(-omega1).*(1./omega1 + 1))./1267650600228229401496703205376 - (238946508779741850518491294873107.*omega1.^3.*exp(-omega1).*(6./omega1 + 15./omega1.^2 + 15./omega1.^3 + 1))./202824096036516704239472512860160 - (15.*pi.*omega1.*exp(-omega1))./2))./omega1.^2));
fun = integral2(temp1.*cos(omega1.*a1.*sin(theta)),0,2.*pi,0,inf)
%%% getting error "Error using integral2 (line 82) , First input argument must be a function handle."
%% I am a beginner and need help in double integration.Thank you in advance!!

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### 採用された回答

Bjorn Gustavsson 2021 年 8 月 13 日
Should look something like this:
fun = integral2(@(theta,omega1) temp1(theta,omega1).*cos(omega1.*a1.*sin(theta)),0,2.*pi,0,inf);
(I'm think I got the ordering of your variables right.) Your temp1 (templ?) variable can be treated as a function pretty much like any other function (built-in, matlab-native of from your toolboxes) except from a computational efficiency viewpoint perhaps.
HTH
##### 2 件のコメント表示非表示 1 件の古いコメント
Bjorn Gustavsson 2021 年 8 月 13 日
My pleasure.
Getting used to function-handles was something that took me a bit longer than the other variable-types - but it is worth the effort.

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R2020b

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