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The formula for integration by parts is given by $$ \int uv'=uv-\int u'v $$ As most of you know. The result is invariant if we use$v=v+c$, instead of $v$ where $c$ is some arbitary constant. $$ \int uv'=u(v+c)-\int u'(v+c) =uv-\int u'v $$ Since $\int{u'c}=uc$ since $c$ is constant. My question is asking for examples where it is more useful to use $v+c$ instead of $v$ when integrating by parts. Concretely examples where $$ \int f(x)\,\mathrm{d}x = (x+c)f(x) - \int(c+x)f'(x)\,\mathrm{d}x $$ Is easier to integrate than $xf'(x)$. One example can be seen here

Evaluating $\int_a^b \arccos\left(x\,/\sqrt{(a+b)x-ab\,}\,\right)\,\mathrm {d}x$ assuming $0<a<b$

Where the integration was simpler/cleaner (perhaps not strictly easier) from choosing $v = x - ab/(a+b)$ instead of plainly $x$.

Are there any other examples of cases where it is simpler to use $v+c$ when integrating by parts? How would one see how to choose a fitting $c$?

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  • $\begingroup$ In practice, there are probably useful answers to this question. From a purely theoretical standpoint, though, it's not really well-posed - because given $v'$ as we are, there isn't one distinguished $v$ to single out from the set of antiderivatives of $v'$. $\endgroup$ – Greg Martin Dec 28 '13 at 22:29
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    $\begingroup$ The Taylor Series and the Euler-Maclaurin Sum Formula are both usually developed using integration by parts with a well chosen constant of integration. $\endgroup$ – robjohn Dec 29 '13 at 0:01
  • $\begingroup$ I first saw the technique here: math.stackexchange.com/a/448974/77151. $\endgroup$ – TooTone Feb 17 '14 at 21:34
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Justifying that $$\begin{align} \int_{0}^{\infty} \frac{\sin x}{x} \, dx &= \int_{0}^{\infty} \int_{0}^{\infty}\sin x \ e^{-xt} \, dt \, dx = \int^{\infty}_{0} \int_{0}^{\infty} \sin x \ e^{-tx} \, dx \, dt \\ &= \int_{0}^{\infty} \frac{1}{1+t^{2}} \, dt = \frac{\pi}{2} \end{align}$$

is not trivial due to the fact that the iterated integral does not converge absolutely.

But if we integrate by parts first and choose the antiderivative $1-\cos x$ for $v$, then

$$ \begin{align} \int_{0}^{\infty} \frac{\sin x}{x} \, dx &= \int_{0}^{\infty}\frac{1- \cos x}{x^{2}} \, dx = \int^{\infty}_{0} \int_{0}^{\infty} (1- \cos x) \, t e^{-xt} \, dt \, dx \\ &= \int_{0}^{\infty} t \int_{0}^{\infty} (1- \cos x) e^{-tx} \, dx \, dt = \int_{0}^{\infty} t \left(\frac{1}{t} - \frac{t}{1+t^{2}} \right) \, dt \\ &= \int_{0}^{\infty} \frac{1}{1+t^{2}} \, dt = \frac{\pi}{2} \, , \end{align}$$

where the justification for switching the order of integration comes from the fact the the integrand is always nonnegative (Tonelli's theorem).

EDIT:

Another example where it's useful to add a constant is for finding a primitive of $ \displaystyle \frac{\log x}{x+a}$.

If we assume that $a > 0$ and choose $\log(x+a) - \log(a) = \log \left(1+ \frac{x}{a} \right)$ for $v$, we get

$$ \begin{align}\int \frac{\log x}{x+a} \, dx &= \log(x) \log \left(1+ \frac{x}{a}\right) - \int \frac{1}{x} \, \log \left(1+ \frac{x}{a} \right) \, dx \\ &= \log(x) \log \left(1+ \frac{x}{a}\right) + \text{Li}_{2} \left(- \frac{x}{a} \right) +C . \tag{1}\end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm

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Here are a couple of simple examples where this applies:

1) $\int x\tan^{-1}x\;dx$, taking $v=\frac{1}{2}x^2+\frac{1}{2}$

2) $\int x\ln(x+1)\;dx$, taking $v=\frac{1}{2}x^2-\frac{1}{2}$

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  • It's very useful to study : $\displaystyle\sum_{k=0}^{+\infty} \frac{\sin \pi\sqrt{n}}{n^\alpha}$ where $\alpha \in \mathbb{R}$

Let $\alpha\leq1$ and denote $(t)=\frac{\sin \pi\sqrt{n}}{n^\alpha}$ By integration by parts $$ \displaystyle\int_{n}^{n+1}f(t)dt-f(n)=\displaystyle\int_{n}^{n+1}(f(t)-f(n))dt $$ $$ =\left[(t-n-1)\bigr(f(t)-f(n)\bigl)\right]_n^{n+1}+\int_{n}^{n+1}(n+1-t)f'(t)dt $$ Thus, $$ \displaystyle\int_{n}^{n+1}f(t)dt-f(n)=\int_{n}^{n+1}(n+1-t)f'(t)dt $$ Then, $$ |\displaystyle\int_{n}^{n+1}f(t)dt-f(n)| \leq \displaystyle\int_{n}^{n+1}\bigr(\frac{\pi}{2t^{\alpha+\frac{1}2}}+\frac{|\alpha|}{t^{\alpha+1}}\bigl)dt=v_n $$ For $\alpha \geq \frac{1}2$ it's easy to conclude, for $\alpha \leq \frac{1}2$ you can use Cauchy criterion.


  • This technique is used to maintain the convergence of integrals

Let $\displaystyle\int_{0}^{+\infty}\frac{sin(t)}{t}dt$, by integration by parts, for $x>0$ $$ \displaystyle\int_{0}^{x}\frac{sin(t)}{t}dt=\left[\frac{1-cos(t)}t\right]_0^{x}+\int_{0}^{x}\frac{1-cos(t)}{t^2}dt $$


Let $I_n=\int_{0}^{n}\bigr(1-\frac{t}n\bigl)^n\ln(t) dt$. Then $I_n=\frac{n}{n+1}\bigr(ln(n)-\displaystyle\sum_{k=1}^{n+1} \frac{1}k\bigl)$.

Using $u=\frac{t}n$, we have $$ I_n=n\int_{0}^{1}\bigr(1-u\bigl)^n\ln(nu) du=nln(n)\int_{0}^{1}\bigr(1-u\bigl)^ndu+n\int_{0}^{1}\bigr(1-u\bigl)^nln(u)du $$ $$ =\frac{-1}{n+1}(1-u)^{n+1}+n\int_{0}^{1}\bigr(1-u\bigl)^nln(u)du $$ Then take $u\mapsto \frac{-1}{n+1}\bigr((1-u)^{n+1}-1\bigl)$ as primitive for $u\mapsto (1-u)^n$. Use integration by parts to conclude and the fact that $x^{n+1}-1=(x-1)(x^n+\cdots+1)$.

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Here are a five other simple examples where this applies:

1) $\int \frac{\ln(x+2)}{(x+1)^{2}}dx$, taking $k=-1.$

2) $\int x\ln(\frac{x-1}{x+1})\;dx$, taking $k=-\frac{1}{2}$

3) $\int x^{2}\ln(1-x^{2}) dx$, taking $k=-\frac{1}{3}$

4) $\int x\ln(x+a)\ln(x-a);dx$, taking $k=-\frac{a^{2}}{2}$

5) $\int \arctan (x^{1/2})\;dx$, taking $k=1$.

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