Articles of integration by parts

Integration by parts of $\cot x$

I attempted to integrate $\cot x$ by parts by taking $u$ as $\csc x$ and $\dfrac{dv}{dx}$ as $\sin x$. Then: $$\int \cot x\,dx = \int \csc x \cos x\,dx \\ = \sin x \csc x – \int- \sin x \csc x \cot x \, dx \\ = \frac{\sin x}{\sin x} + \int \frac{\sin x}{\sin x}\cot […]

Integration by Parts? – Variable Manipulation

$$\int x^3f”(x^2)\,\mathrm{d}x$$ Solve using Integration by Parts. \begin{align} u&=x^3\qquad\mathrm{d}v=f”(x^2) \\ \mathrm{d}u&=3x^2\qquad v=f'(x^2) \\ &=x^3f'(x)-\int f'(x^2)3x^2 \\ u&=3x^2\quad\mathrm{d}v=f'(x^2) \\ \mathrm{d}u&=6x \qquad v=f(x^2) \\ &=x^3f'(x^2)-[3x^2f(x^2)-\int f(x^2)6x] \end{align} No clue what do from here as the correct answer is: $$\frac{1}{2}(x^2f'(x^2)-f(x^2))+C$$ Can you guys think of anything? I appreciate the help. 🙂

Can you integrate by parts with one integral inside another?

From the definition of the Laplace transform: $$\mathcal{L}[f(t)]\equiv \int_{t=0}^{\infty}f(t)e^{-st}\mathrm{d}t$$ where $s \in \mathbb{R^+}$. $$\mathcal{L}\left[\int_{u=0}^{t}f(u)\mathrm{d}u\right]=\int_{t=0}^{\infty}e^{-st}\mathrm{d}t\int_{u=0}^{t}f(u)\mathrm{d}u$$ $$=\int_{t=0}^{\infty}e^{-st}\mathrm{d}t\int_{u=0}^{t} f(u)\mathrm{d}u=\int_{u=0}^{t} f(u) \mathrm{d}u\int_{t=0}^{\infty}e^{-st}\mathrm{d}t$$ $$\left[-\cfrac{1}{s}e^{-st} \int_{u=0}^{t} f(u)\mathrm{d}u \right]_{\color{red}{0}}^{\color{red}{\infty}}+\int_{\color{red}{0}}^{\color{red}{\infty}}\cfrac{1}{s}e^{-st}f(\color{blue}{t})\mathrm{d}\color{blue}{t}$$ $$=\cfrac{1}{s}\mathcal{L}\left[\color{#180}{f}\right]$$ Assuming that on page 486 in this book it was done by parts I have $\mathbf{3}$ questions about the calculation above: $\mathbf{\color{red}{1)}}$ For the $2$ sets of limits marked red above for […]

Deriving the Normalization formula for Associated Legendre functions: Stage $4$ of $4$

The question that follows is the final stage of the previous $3$ stages found here: Stage 1, Stage 2 and Stage 3 which are needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{2}$$ are a set of Associated Legendre functions on $[−1, 1]$. […]

Computing the Fourier series of $f(\theta)=(\pi-\theta)^2/4$ for $0\leq\theta\leq{2\pi}$

According to p. 36 of Stein and Shakarchi’s Fourier Analysis, the Fourier series is $\frac{\pi^2}{12} + \sum_{n=1}^{\infty} \frac {\cos{n\theta}}{n^2}$. But to calculate the Fourier coefficients, I computed $\int_{-\pi}^{\pi}\frac{(\pi-\theta)^2}{4}e^{-in\theta}d\theta = \frac {1} {4} \Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{2}{in}\int_{-\pi}^{\pi}e^{-in\theta}(\pi-\theta)d\theta\Bigg]=\frac {1} {4}\Bigg[\frac{4\pi^2e^{in\pi}}{in}-\frac{4\pi{e^{in\pi}}+2e^{-in\pi}-2e^{in\pi}}{n^2}\Bigg]=\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}$ This means the Fourier series is $\sum_{n=-\infty}^{\infty}\frac{\pi^2ne^{in\pi}+\sin{n\pi}}{in^2}e^{in\theta}=\sum_{n=-\infty}^{\infty}\frac{\pi^2e^{in(\pi+\theta)}}{in}+\frac{e^{in(\theta-\pi)}}{2n^2}$, but I can’t seem to get this to equal that simple thing from […]

Use integration by parts to prove $ \int^\infty_0 x^ne^{-x} dx=n!$

$ \int^\infty_0 (x^n)(e^{-x}) dx $ and show that it is equal to $(n!)$ ? I know that if you differentiate $x^n$ infinitely you get $n!$ but I don’t know how to prove?

Methods for choosing $u$ and $dv$ when integrating by parts?

When doing integration by parts, how do you know which part should be $u$ ? For example, For the following: $$\int x^2e^xdx$$ $u = x^2$? However for: $$\int \sqrt{x}\ln xdx$$ $u = \ln x$? Is there a rule for which part should be $u$ ? As this is confusing.

Infinite Integration by Parts

A while back, I was explaining to a friend a method to compute the indefinite integral of $\ln(x)$, by taking $\int \ln(x) \, dx$ and setting $u = \ln(x), du = \frac{1}{x} \, dx, dv = 1 \, dx, v = x$, then proceeding with integration by parts. However, I was wondering if this method […]