7
$\begingroup$

Usually, integration by parts only works one way. For example, evaluating $\int xe^x\,dx$ can only be done by differentiating $x$ and integrating $e^x$, but not the other way round, since $\int x^2e^x\,dx$ makes it more difficult than the original integral.

However, the integral $$\int x\ln x\,dx$$ can be evaluated using integration by parts both ways:

$$\int x\ln x\,dx=x\int \ln x\,dx - \int\left(x'\int\ln x\,dx\right)\,dx=x^2(\ln x-1)-\int x(\ln x-1)\,dx$$ and (directly), $$\int x\ln x\,dx=\ln x\int x\,dx-\int\left(\ln'x\int x\,dx\right)\,dx=\frac{x^2}2\ln x-\int\frac x2\,dx.$$

There are also trigonometric integrals that can do this. For instance, an integral consisting of $\sin$ and $\cos$, as $\int \sin = -\cos$ and $\int \cos = \sin$: $$\int \sin2x\cos3x\,dx.$$

@JamesArathoon has provided this one as well: $$\int x\arctan x\,dx$$

What are some other examples of integrals that have this property as well?

I believe that such integrals are rather rare, so I don't want 'similar' integrals like multiplying by a constant or adding $1$ to the integrand.

Of course, integrands of forms other than $xf(x)$ would be even better (and more challenging :)

$\endgroup$
16
  • 2
    $\begingroup$ You can not write $x$ in front of your integral! $\endgroup$ – Dr. Sonnhard Graubner Jul 4 '18 at 19:39
  • 2
    $\begingroup$ @Dr.SonnhardGraubner Look again. I am using integration by parts. Did you forget to look at the term after that? $\endgroup$ – TheSimpliFire Jul 4 '18 at 19:40
  • 3
    $\begingroup$ @Sonnhard: Let $u=x,dv=\ln x\,dx.$ Then $$\int u\,dv=uv-\int v\,du,$$ no? That's what the OP has written. $\endgroup$ – Cameron Buie Jul 4 '18 at 19:45
  • 2
    $\begingroup$ @Dr.SonnhardGraubner Application to find antiderivatives $\endgroup$ – Jon Jul 4 '18 at 19:45
  • 2
    $\begingroup$ @Nick Perhaps. It would be great if there were. Congrats for persevering with the election btw :) $\endgroup$ – TheSimpliFire Aug 5 '18 at 15:11
4
+50
$\begingroup$

So we wish for $f(x)$ and $g(x)$ such that we can evaluate $\int f(x) g(x) dx$ using integration by parts in two ways, either by A:

$$\int f(x) g(x) dx = g(x)\int f(x) dx - \int (g'(x) \int f(x) dx ) dx $$

... or by B:

$$\int f(x) g(x) dx = f(x)\int g(x) dx - \int (f'(x) \int g(x) dx ) dx $$

In the following list, (i) to (iii) are put forward by yourself, (iv) and (v) are trivial examples from the comments, and (vi) and (vii) are new.

(i) $f(x) = x$ and $g(x) = \mathrm{ln}(x)$

(ii) $f(x) = \mathrm{sin}(mx) $ and $g(x) =\mathrm{cos}(nx) $ for $n \neq m$

(iii) $f(x) = x $ and $g(x) = \mathrm{arctan}(x) $

(iv) $f(x) = x^n $ and $g(x) = x^m $

(v) $f(x) = e^{nx} $ and $g(x) = e^{mx} $

(vi) $f(x) = e^{nx} $ and $g(x) = \mathrm{sin}(mx) $

(vii) $f(x) = x $ and $g(x) = \sqrt{1+x^2}$

Other examples can be made by switching around some trig and hyperbolic functions (e.g. $\mathrm{sin}$ and $\mathrm{cos}$ for $\mathrm{sinh}$ and $\mathrm{cosh}$ in (ii) or $\mathrm{arctan}$ for $\mathrm{arcsin}$ in (iii)).

It sometimes seems the case that one of the directions A or B is much easier than the other. I think this is true of (i), (iii) and (vii) (and I have written them so that A is easier than B).

For all of the examples except for the trivial (iv) and (v), at least one of A and B require the use of the "indirect method", where the integrand $\int f(x)g(x) dx$ is found again on the RHS and the expression is rearranged to get $\alpha \int f(x)g(x) dx$ on the LHS with $\alpha \neq 1$.

So far I cannot spot any other interesting similarities between these examples... Perhaps others can, or could add to the list?

Edit: another example

(viii) $f(x)=x$ and $g(x)=\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int^{x}_0 e^{-t^2} dt$

(We can show that $ \int \mathrm{erf}(x) dx = x\mathrm{erf}(x) + \frac{e^{-x^2}}{\sqrt{\pi}} + C $ by another application of integration by parts, and $\frac{d}{dx} \mathrm{erf}(x)= \frac{2}{\sqrt{\pi}} e^{-x^2} $ fairly trivially.)

Interestingly, evaluating (viii) using B requires use of the “indirect method” again.

I also tried using $f(x)=x$ and $g(x)=\Gamma (x)$, but the integral is undefined.

Edit: proof of equivalence of methods A and B when $f(x)=x$

@farruhota suggests in the comments that (i) might be based on some sort of circular argument, as $\int \mathrm{ln}(x) dx $ is often itself evaluated using integration by parts. The same could be said of $\int \mathrm{arctan}(x) dx$ in (iii) and of $\int \mathrm{erf}(x) dx$ in (viii). I now believe that this is true and is very much linked to why the "indirect method" keeps cropping up in method B. I offer the following proof.

Let $f(x)=x$. Then method A goes:

$$ \begin{align} I(x) &:= \int^x_0 tg(t) dt \\ &= \frac{1}{2}x^2 g(x)-\frac{1}{2} \int^x_0 t^2 g'(t) dt \qquad \qquad (1) \end{align} $$

... and the integral $\int^x_0 t^2 g'(t) dt$ must then be evaluated.

(Note that I am now using limits of integration in order to make my workings more rigorous.)

Method B begins instead:

$$ \begin{align} I(x) &= \int^x_0 tg(t)dt \\ &= x \int^x_0 g(t)dt - \int^x_0 \left( \int^t_0 g(s)ds \right) dt \qquad \qquad (2) \end{align} $$

Now I'll use integration by parts again in order to evaluate $\int^x_0 g(t)dt$ and $\int^t_0 g(s)ds$, so that $(2)$ follows on as:

$$ \begin{align} I(x) &= x \left( xg(x)-\int^x_0 tg'(t)dt \right) - \int^x_0 \left( tg(t) - \int^t_0 sg'(s)ds \right) dt \\ &= x^2g(x) - x \int^x_0 tg'(t)dt - I(x) + \int^x_0 \int^t_0 sg'(s)ds \, dt \end{align} $$

Now here's where the "indirect method" comes in. We can collect $I(x)$ onto the LHS and divide by $2$ so that:

$$ I(x) = \frac{1}{2} x^2g(x) - \frac{1}{2} \underbrace{ \left( x \int^x_0 tg'(t)dt - \int^x_0 \int^t_0 sg'(s)ds \, dt \right) }_{=:K(x)} \qquad \qquad (3) $$

Focus now on $K(x)$. Write $h(t):=tg'(t)$ so that:

$$ \begin{align} K(x) &= x \int^x_0 h(t)dt - \int^x_0 \int^t_0 h(s)ds \, dt \\ &= \int^x_0 th(t) dt \\ &= \int^x_0 t^2g'(t) \end{align} $$

... where I've used integration by parts "backwards" to get from line 1 to line 2.

Putting this back into $(3)$, we get:

$$ I(x)= \frac{1}{2}x^2 g(x)-\frac{1}{2} \int^x_0 t^2 g'(t) dt \qquad \qquad (4)$$

... which is exactly the same as method A at $(1)$!

Both methods A and B boil down to whether we can evaluate $\int^x_0 t^2 g'(t) dt$. I hence offer the following as a (partial) answer to the question.

If $f(x)=x$ and $g(x)$ is such that we can evaluate $\int x^2 g'(x) dx$ then $\int f(x) g(x) dx$ can be integrated by parts "both ways".

Claim: A and B are always equivalent

Now, you suggest finding examples where $f(x)$ is not just equal to $x$.

If equations $(2)$ to $(4)$ are worked through with a general $f(x)$ instead of with $f(x)=x$ then the following can be obtained:

$$ \int^x_0 f(t)g(t)dt = xf(x)g(x) - \int^x_0 tf'(t)g(t)dt - \int^x_0 tf(t)g'(t)dt \qquad (5) $$

Note that $f(x)=x$ worked nicely because $f(x)=kx \iff tf'(t)=f(t)$ and so the second term on the RHS of $(5)$ becomes $I(x)$.

Also note that $(5)$ is symmetric in $f$ and $g$. It can be obtained by beginning with either route A or B. I would hence conclude that routes A and B are always equivalent:

route A can be used iff B can be used, as both can be manipulated into $(5)$

(However, one of these routes might be much longer than the other.)

$\endgroup$
4
  • $\begingroup$ Very interesting answer! Do you mean that the property holds if $\int x^2g'(x)\,dx$ can be evaluated without integration by parts? $\endgroup$ – TheSimpliFire Aug 8 '18 at 7:46
  • $\begingroup$ I don't think it matters. In examples (i), (iii) and (vii), integration by parts is not needed again to evaluate $\int x^2g'(x)dx$ (instead, substitutions are required for (iii) and (vii) while (i) is trivial). However, another nice application of integration by parts is needed to evaluate $\int x^2g'(x)dx = \frac{2}{\sqrt{\pi}}\int x^2e^{-x^2}dx$ in (viii) (with one of the "parts" being $x$ and the other being $xe^{-x^2}$ and the solution being in terms of $\mathrm{erf}(x)$). $\endgroup$ – Malkin Aug 8 '18 at 8:59
  • $\begingroup$ But what about $g(x)=e^x$? Since $\int x^2e^x\,dx$ can be evaluated using IBP, does that mean it satisfies the property? $\endgroup$ – TheSimpliFire Aug 8 '18 at 9:27
  • $\begingroup$ I think this is a difficulty in working with such a loose definition. What do you think? It seems like evaluating $\int x^2e^xdx$ using IBP (with one part $x^2$ and one part $e^x$) is "cheating" because it's like taking one step forward and two steps back. But evaluating $\frac{2}{\sqrt{\pi}}\int x^2e^{-x^2}dx$ (with one part $x$ and one part $xe^{-x^2}$) seems fine because because you're not going back on yourself. $\endgroup$ – Malkin Aug 8 '18 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.