Trying to use the integration by parts trick with an extra term

the question is as follows.

$\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$ can you solve this with integration by parts trick? it looks odd.. you obviously don't wanna integrate the y term which means u wanna derivate it then integrate in the next step with the other piece to get the stuff in the middle to be your integral but i can't seem to do it correctly.


A more clear explanation of what hes doing below is ( which i may of fudged up )

let F(x)=$\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$ then derivative with respect to x is $F^{'}(x)$= (by chain rule) $2*(y^{-1} e^{xy} \cos (y)) + I(x)$ evaluated at $2x$ because the derivative of 1=0 $2*(2x^{-1} e^{2x^{2}} \cos (2x))$ =$(x^{-1} e^{2x^{2}} \cos (2x))$ + I(x) now I(x) is just the derivative of the function with respect x holding y constant so $\int^{2x}_{1}y*y^{-1} e^{xy} \cos (y)$ the y comes form the derivative of $e^{xy}$

now we have

$\int^{2x}_{1} e^{xy} \cos (y)dy$ by integration by parts integrate $\cos(y)$ derivate $e^{xy}$ we get $e^{xy} \sin (y)$ - $\int^{2x}_{1} xe^{xy} \sin (y)dy$ again by parts we get $e^{xy} \sin (y)$- $(- xe^{xy} \cos (y)dy)$ - $(x^{2} \int^{2x}_{1} e^{xy} \cos (y)dy)$ the second integral is the same as the first so I(x)= $e^{xy} \sin (y)$- $(- xe^{xy} \cos (y)dy)$ - $(x^{2} I(x))$

thus I(x) =$(e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2}) |^{2x}_{1}$

thus $F^{'}(x)= (x^{-1} e^{2x^{2}} \cos (2x)) + (e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2})|^{2x}_{1}$

Thus $F(x)= \int [ (x^{-1} e^{2x^{2}} \cos (2x)) + (e^{xy} \sin (y)+ xe^{xy} \cos (y))/(1+x^{2})|^{2x}_{1})] dx$

  • $\begingroup$ I think integration by parts deserves to be called a theorem. $\endgroup$ – Julien Mar 22 '13 at 4:42
  • $\begingroup$ haha its not the theorem i was really inquiring about but the I trick that is written below $\endgroup$ – Faust Mar 22 '13 at 4:45

Let $I(x)=\int^{2x}_{1} y^{-1} e^{xy} \cos (y) dy$. We have $I(1)=0, I'(x)=2y^{-1} e^{xy} \cos (y)+\int^{2x}_{1} e^{xy} \cos (y) dy$ The indefinite integral yields to two integrations by parts, differentiating the $\cos y$ and then the $\sin y$ that results, returning the same integral with a factor $-x^2$ so we have $\int^{2x}_{1} e^{xy} \cos (y) dy=\frac{e^{xy}(x \cos y + sin y)}{1+x^2}|_1^{2x}$ Now integrate with respect to $x$

  • $\begingroup$ i actually understand that but there no way i can integrate that with respect to x $\endgroup$ – Faust Mar 22 '13 at 4:51
  • $\begingroup$ Did you think by any chance that the bounds of integration of $I(0)$ were equal when $x=1$? According to WA, $I(1)$ is small, but nonzero. I think you want to do the trick with $I(1/2)$. $\endgroup$ – Julien Mar 22 '13 at 4:56
  • $\begingroup$ he took the derivative of the integral with respect to x making one of the bound 0 * ( stuff) and the other bound 2*(stuff) then used the a simple trick to integrate in y now wants to integrate in x then evaluate the function ( i think) $\endgroup$ – Faust Mar 22 '13 at 4:58

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