# Examples of integration by parts that work both ways

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 :)

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

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.)

\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.)

• Very interesting answer! Do you mean that the property holds if $\int x^2g'(x)\,dx$ can be evaluated without integration by parts? – TheSimpliFire Aug 8 '18 at 7:46
• 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 $\mathrm{erf}(x)$). – Malkin Aug 8 '18 at 8:59
• 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? – TheSimpliFire Aug 8 '18 at 9:27
• 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. – Malkin Aug 8 '18 at 10:45