Post History
#4: Post edited
by
Michael Hardy
·
2024-07-17T16:26:47Z (4 months ago)
Such an example is $\sin(1/u)$ for $0 < u < 1. $- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
\text{Then } & \frac{du}u = -\frac{dx}x.- \end{align}
- For $1\le a < b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
- $$
\int_{1/b}^{1/a} \sin \frac1u~du = -\int_b^a \frac{\sin x}x~dx = \int_a^b \frac{\sin x} x~dx.- $$
- Such an example is $\dfrac{\sin(1/u)} u$ for $0 < u < 1. $
- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
- \text{Then } & du = -\frac{dx}{x^2}.
- \end{align}
- For $1\le a < b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
- $$
- \int_{1/b}^{1/a} \frac{\sin(1/u)} u ~du = -\int_b^a \frac{\sin x}x~dx = \int_a^b \frac{\sin x} x~dx.
- $$
#3: Post edited
by
Michael Hardy
·
2024-07-17T16:23:36Z (4 months ago)
- Such an example is $\sin(1/u)$ for $0 < u < 1. $
- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
- \text{Then } & \frac{du}u = -\frac{dx}x.
- \end{align}
- For $1\le a < b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
- $$
\int_{1/b}^{1/a} \sin \frac1u\, du = -\int_b^a \frac{\sin x}x\,dx = \int_a^b \frac{\sin x} x \, dx.- $$
- Such an example is $\sin(1/u)$ for $0 < u < 1. $
- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
- \text{Then } & \frac{du}u = -\frac{dx}x.
- \end{align}
- For $1\le a < b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
- $$
- \int_{1/b}^{1/a} \sin \frac1u~du = -\int_b^a \frac{\sin x}x~dx = \int_a^b \frac{\sin x} x~dx.
- $$
#2: Post edited
by
Michael Hardy
·
2024-07-17T16:22:58Z (4 months ago)
Such an example is $\sin(1/u)$ for $0<u<1.$- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
- \text{Then } & \frac{du}u = -\frac{dx}x.
- \end{align}
For $1\le a<b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so- $$
int_{1/b}^{1/a} \sin \frac1u\, du = -\int_b^a \frac{\sin x}x\,dx = \int_a^b \frac{\sin x} x \, dx.- $$
- Such an example is $\sin(1/u)$ for $0 < u < 1. $
- \begin{align}
- \text{Let } & u = \frac1x. \\[6pt]
- \text{Then } & \frac{du}u = -\frac{dx}x.
- \end{align}
- For $1\le a < b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
- $$
- \int_{1/b}^{1/a} \sin \frac1u\, du = -\int_b^a \frac{\sin x}x\,dx = \int_a^b \frac{\sin x} x \, dx.
- $$
#1: Initial revision
by
Michael Hardy
·
2024-07-17T16:20:13Z (4 months ago)
Such an example is $\sin(1/u)$ for $0<u<1.$
\begin{align}
\text{Let } & u = \frac1x. \\[6pt]
\text{Then } & \frac{du}u = -\frac{dx}x.
\end{align}
For $1\le a<b,$ as $x$ goes from $a$ to $b,$ $u$ goes from $1/a$ to $1/b$, so
$$
int_{1/b}^{1/a} \sin \frac1u\, du = -\int_b^a \frac{\sin x}x\,dx = \int_a^b \frac{\sin x} x \, dx.
$$