I know that some integrals on solving by parts end up with the same integral on the right side, and then the integral is assumed to be $I$ or some variable, and then linearly solved. For example, $$I = \int e^x \cos x \ dx$$ $$u = e^x,\ dv = \cos x \ dx, \ du = e^x \ dx,\ v = \sin x$$ $$\int u dv = uv - \int v \ du$$ $$I = e^x \sin x - \int e^x \sin x \ dx$$ Similarly, $$u = \cos x,\ dv = e^x \ dx, \ du = -\sin x \ dx,\ v = e^x$$ $$\int u dv = uv - \int v \ du$$ $$I = e^x \cos x - \int - \ e^x \sin x \ dx$$ $$I = e^x \cos x + \int e^x \sin x \ dx$$ Adding the two values of $I$ and dividing by 2, we obtain $$I = \frac{e^x(\sin x + \cos x)}{2} + C$$ But when I tried this for $1/x$, $$I = \int \frac{dx}{x}$$ $$u = \frac{1}{x},\ dv = dx, \ du = -\frac{dx}{x^2},\ v = x$$ $$\int u dv = uv - \int v \ du$$ $$I = x\frac{1}{x} - \int - x \frac{1}{x^2} \ dx$$ $$I = 1 + \int \frac{1}{x} \ dx$$ $$I = 1 + I$$ What is happening here?
-
Integration gives family of curves by writing a $c$ we get a specified curve here i think $c=1$ – Archis Welankar Feb 06 '16 at 08:32
-
Related http://math.stackexchange.com/questions/424854/1-0-by-integration-by-parts-of-tanx – GoodDeeds Feb 06 '16 at 09:04
2 Answers
Given an interval $I\subset{\mathbb R}$ and a function $$f:\quad I\to{\mathbb R}, \qquad x\mapsto f(x)\ ,$$ the indefinite integral $\int f(x)\>dx$ is a family of functions on $I$, usually denoted by "$\>F(x)+C\>$". For the purpose of this answer let's write $\langle F(x)\rangle$, or simply $\langle F\rangle$, instead .
It is a little exercise to prove that such families can be added and scaled according to $$\langle F\rangle+\langle G\rangle:=\langle F+G\rangle,\qquad \lambda\langle F\rangle:=\langle \lambda F\rangle\ ,$$ so that the set of all these families forms a real vector space. This then allows for arguments of the kind used in the calculation of integrals $\int e^x\>\cos x\>dx$, and similar.
- 226,825
-
Well put.In other words $\int u dv=u v-\int v du$ actually says that if $dF(x)/dx=u(x)v'(x)$ and $dG(x)/dx=v(x)u'(x)$ then $ F(x)- [u(x)v(x)-G(x)]$ is a constant. An indefinite integral $\int u dv$ is not a number, nor a function, but an infinite family of functions, each differing from another by a constant. So the integration-by-parts equation states that the family on the LHS is equal to the family on the RHS. We can do some algebraic operations on such families as if they were numbers but the family $F$ and the family $F+1$ are the same family. – DanielWainfleet Feb 06 '16 at 11:23
The only thing you've done wrong with the integration of $\dfrac 1x$ is to assume at the last step that the constant of integration was $0$. There are constants of integration floating around in the first example ($\int e^x\cos x$) attached to the indefinite integrals, but they all get merged together into the final ${}+C$.
Let $\displaystyle\int\dfrac1x~dx = F(x)+C$. Your deduction $I=1+I$ is $F(x)+C_1=1+F(x)+C_2$ or $0=1+C$ where $C=C_2-C_1$.
In the other examples of integration by parts you've seen using this technique, the integrals end up as $kI=F(x)+C$ for some $k\ne0$, so $I=\dfrac1kF(x)+C'$. Unfortunately in your case, $k=0$ and we can't deduce much about $I$ from $0I=0$.
- 3,041