Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals?

As an example of something that might be obscure, the formula for "general integration by parts " for $n$ functions $f_j, \ j = 1,\cdots,n$ is given by $$ \int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}} $$ which is not necessarily useful nor difficult to derive, but is interesting nonetheless.

So out of curiosity, are there any crazy unknown symbolic integration techniques?

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    $\begingroup$ This seems a bit relevant: web.williams.edu/Mathematics/lg5/Feynman.pdf $\endgroup$ – skrub Sep 23 '14 at 0:41
  • $\begingroup$ Sometimes converting integrals into laplace transform type problems is useful $\endgroup$ – ClassicStyle Sep 23 '14 at 4:59
  • $\begingroup$ Using power series expansions of the integrand can be funny $\endgroup$ – Avitus Sep 23 '14 at 11:04
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    $\begingroup$ Irresistible Integrals by Boros and Moll has already been cited, but here's a great book hot off the presses: Inside Interesting Integrals by Paul J. Nahin (Springer, 2015). Nahin is a retired electrical/computer engineer who has written some great books that acknowledge and show concern for a mathematician's desire for rigor. $\endgroup$ – PolyaPal Sep 23 '14 at 21:32
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    $\begingroup$ Feynnman's method of integration is also good. $\endgroup$ – Aditya Kumar Dec 3 '15 at 4:47

15 Answers 15


Here are a few. The first one is included because it's not very well known and is not general, though the ones that follow are very general and very useful.

  • A great but not very well known way to find the primitive of $f^{-1}$ in terms of the primitive of $f$, $F$, is (very easy to prove: just differentiate both sides and use the chain rule): $$ \int f^{-1}(x)\, dx = x \cdot f^{-1}(x)-(F \circ f^{-1})(x)+C. $$


$$ \begin{aligned} \displaystyle \int \arcsin(x)\, dx &= x \cdot \arcsin(x)- (-\cos\circ \arcsin)(x)+C \\ &=x \cdot \arcsin(x)+\sqrt{1-x^2}+C. \end{aligned} $$

$$ \begin{aligned} \int \log(x)\, dx &= x \cdot \log(x)-(\exp \circ \log)(x) + C \\ &= x \cdot \left( \log(x)-1 \right) + C. \end{aligned} $$

  • This one is more well known, and extremely powerful, it's called differentiating under the integral sign. It requires ingenuity most of the time to know when to apply, and how to apply it, but that only makes it more interesting. The technique uses the simple fact that $$ \frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y. $$


We want to calculate the integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$. To do that, we unintuitively consider the more complicated integral $\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$ instead.

Let $$ I(t)=\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx,$$ then $$ I'(t)=-\int_{0}^{\infty} e^{-tx} \sin(x) dx=\frac{e^{-t x} (t \sin (x)+\cos (x))}{t^2+1}\bigg|_0^{\infty}=\frac{-1}{1+t^2}.$$

Since both $I(t)$ and $-\arctan(t)$ are primitives of $\frac{-1}{1+t^2}$, they must differ only by a constant, so that $I(t)+\arctan(t)=C$. Let $t\to \infty$, then $I(t) \to 0$ and $-\arctan(t) \to -\pi/2$, and hence $C=\pi/2$, and $I(t)=\frac{\pi}{2}-\arctan(t)$.

Finally, $$ \int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0) = \frac{\pi}{2}-\arctan(0) = \boxed{\frac{\pi}{2}}. $$

  • This one is probably the most commonly used "advanced integration technique", and for good reasons. It's referred to as the "residue theorem" and it states that if $\gamma$ is a counterclockwise simple closed curve, then $\displaystyle \int_\gamma f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( f, a_k )$ . It will be difficult for you to understand this one without knowledge in complex analysis, but you can get the gist of it with the wiki article. Example:

    We want to compute $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$. The poles of our function $f(z)=\frac{x^2}{1+x^4}$ in the upper half plane are $a_1=e^{i \frac{\pi}{4}}$ and $a_2=e^{i \frac{3\pi}{4}}$. The residues of our function at those points are $$\operatorname{Res}(f,a_1)=\lim_{z\to a_1} (z-a_1)f(z)=\frac{e^{i \frac{-\pi}{4}}}{4},$$ and $$\operatorname{Res}(f,a_2)=\lim_{z\to a_2} (z-a_2)f(z)=\frac{e^{i \frac{-3\pi}{4}}}{4}.$$ Let $\gamma$ be the closed path around the boundary of the semicircle of radius $R>1$ on the upper half plane, traversed in the counter-clockwise direction. Then the residue theorem gives us ${1 \over 2\pi i} \int_\gamma f(z)\,dz=\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_2)={1 \over 4}\left({1-i \over \sqrt{2}}+{-1-i \over \sqrt{2}}\right)={-i \over 2 \sqrt{2}}$ and $ \int_\gamma f(z)\,dz= {\pi \over \sqrt{2}}$. Now, by the definition of $\gamma$, we have: $$\int_\gamma f(z)\,dz = \int_{-R}^R \frac{x^2}{1+x^4} dx + \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz = {\pi \over \sqrt{2}}.$$ For the integral on the semicircle $$ \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz, $$ we have $$ \begin{aligned} \left| \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \right| &\leq \int_0^\pi \left| {i (R e^{it})^3 \over 1+(R e^{it})^4} \right| dz \\ &\leq \int_0^\pi {R^3 \over R^4-1} dz={\pi R^3 \over R^4-1}. \end{aligned} $$ Hence, as $R\to \infty$, we have ${\pi R^3 \over R^4-1} \to 0$, and hence $\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \to 0$. Finally, $$ \begin{aligned} \int_{-\infty}^\infty \frac{x^2}{1+x^4} dx &= \lim_{R\to \infty} \int_{-R}^R \frac{x^2}{1+x^4} dx \\ &= \lim_{R\to \infty} {\pi \over \sqrt{2}}-\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz =\boxed{{\pi \over \sqrt{2}}}. \end{aligned} $$

  • My final "technique" is the use of the mean value property for complex analytic functions, or Cauchy's integral formula in other words: $$ \begin{aligned} f(a) &= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-a}\, dz \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} f\left(a+e^{ix}\right) dx. \end{aligned} $$


We want to compute the very messy looking integral $\int_0^{2\pi} \cos (\cos (x)+1) \cosh (\sin (x)) dx$. We first notice that $$ \begin{aligned} &\hphantom{=} \cos [\cos (x)+1] \cosh [\sin (x)] \\ &=\Re\left\{ \cos [\cos (x)+1] \cosh [\sin (x)] -i\sin [\cos (x)+1] \sinh [\sin (x)] \right\} \\ &= \Re \left[ \cos \left( 1+e^{i x} \right) \right]. \end{aligned} $$ Then, we have $$ \begin{aligned} \int_0^{2\pi} \cos [\cos (x)+1] \cosh [\sin (x)] dx &= \int_0^{2\pi} \Re \left[ \cos \left( 1+e^{i x} \right) \right] dx \\ &= \Re \left[ \int_0^{2\pi} \cos \left( 1+e^{i x} \right) dx \right] \\ &= \Re \left( \cos(1) \cdot 2 \pi \right)= \boxed{2 \pi \cos(1)}. \end{aligned} $$

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    $\begingroup$ +1, how about monte-carlo integration or sth along these lines as well, this could go (and applied) a long way and provide results not easily derive by other methods (if derived at all)? Plus how about some integration tricks used in Quantum Mechanics (eg Feynman had a couple of tricks) $\endgroup$ – Nikos M. Sep 23 '14 at 22:37
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    $\begingroup$ @NikosM. Monte Carlo does numerical integration, not symbolic. $\endgroup$ – Jack M Sep 25 '14 at 8:37
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    $\begingroup$ @JackM, of course, however it relates a (definite) integral to random variable, i think this counts as an advanced technique, which can derive results not easily derived otherwise. Mpreove the comment was also about techniques used mostly in physics (e.g differentiation by a paramater, some of Feynman's tricks etc..) $\endgroup$ – Nikos M. Sep 25 '14 at 23:55
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    $\begingroup$ @NikosM. I assumed that OP wanted analytic solutions to his integrals, not numerical approximations. Although, feel free to make a post about it in this thread, numerical analysis is of course a very interesting subject, just not sure if it's what the OP wants :-). $\endgroup$ – Fujoyaki Sep 25 '14 at 23:59

You can do integration by inverting the matrix representation of the differentiation operator with respect to a clever choice of a basis and then apply the inverse of the operator to function you wish to integrate.

For example, consider the basis $\mathcal{B} = \{e^{ax}\cos bx, e^{ax}\sin bx \}$. Differentiating with respect to $x$ gives \begin{align*} \frac{d}{dx}e^{ax} \cos bx &= ae^{ax} \cos bx - be^{ax} \sin bx\\ \frac{d}{dx} e^{ax} \sin bx &= ae^{ax} \sin bx + be^{ax} \cos bx \end{align*}

and the matrix representation of the linear operator is

$$T = \begin{bmatrix} a & b\\ -b & a \end{bmatrix}$$

To then solve something like $\int e^{ax}\cos bx\operatorname{d}\!x$, this is equivalent to calculating

$$T^{-1}\begin{bmatrix} 1\\ 0 \end{bmatrix}_{\mathcal{B}} = \frac{1}{a^{2} + b^{2}}\begin{bmatrix} a\\ b \end{bmatrix}_{\mathcal{B}}.$$

That is,

$$\int e^{ax}\cos bx\operatorname{d}\!x = \frac{a}{a^{2}+b^{2}}e^{ax}\cos bx + \frac{b}{a^{2} + b^{2}}e^{ax}\sin bx$$

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    $\begingroup$ This is actually really cool! I didn't quite understand it when you first posted it, but now I see just how awesome it truly is! $\endgroup$ – user3002473 Dec 17 '14 at 2:41

For a really advanced technique, you may want to read about Risch's algorithm for indefinite integration, which is implemented in the major symbolic mathematics programs.

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    $\begingroup$ I assume this is not something to do by hand? $\endgroup$ – Simply Beautiful Art Feb 17 '17 at 21:11
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    $\begingroup$ @SimplyBeautifulArt define "by hand". There are also multiple levels of complexity within the Risch Algorithm... We can get more basic results pretty easily, but even most major Mathematical softwares don't implement the whole Risch Algorithm $\endgroup$ – Brevan Ellefsen Feb 19 '17 at 18:46

Another option is converting the value under the integral to a summation. For example,

$$ \int{\frac{1}{1 + x^2}dx} = \int\sum_{i = 0}^\infty{(-1)^ix^{2i}}dx = \sum_{i = 0}^\infty(-1)^i\int{x^{2i}}dx = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i+1}}{2i + 1}.$$

You might then make use of the fact that,

$$\sum_{i = 0}^\infty \frac{(-1)^ix^{2i+1}}{2i + 1} = \tan^{-1}{x}.$$

Of course, you need to be familiar with many different series, which comes with practise. In fact, most derivations of $\arctan(x)$ as a series actually use the method I just used. However, it still serves as an example of the technique.

Another example of this comes through the Riemann zeta function:

Let $u=kx$,

$$\begin{align}\int_0^\infty\frac{x^s}{e^x-1}\ dx&=\int_0^\infty x^se^{-x}\left(\frac1{1-e^{-x}}\right)\ dx\\&=\int_0^\infty x^se^{-x}\sum_{k=0}^\infty e^{-kx}\ dx\\&=\sum_{k=1}^\infty\int_0^\infty x^se^{-kx}\ dx\\&=\sum_{k=1}^\infty\frac1{k^{s+1}}\int_0^\infty u^se^{-u}\ du\\&=\zeta(s+1)\Gamma(s+1)\end{align}$$

A beautiful non-trivial example of expansion and solving through series.

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    $\begingroup$ What's the name of the theorem that alows you to interchange sum and integral like that? $\endgroup$ – zerosofthezeta Sep 24 '14 at 4:16
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    $\begingroup$ See here. Of course, there's always other ways of showing it. $\endgroup$ – Chantry Cargill Sep 24 '14 at 4:49
  • $\begingroup$ @ChantryCargill I've given you a much less trivial example of how this can be useful ;) In general, you can try looking at the following integral: $$\int_0^\infty\frac{x^s}{e^x+a}\ dx$$ $\endgroup$ – Simply Beautiful Art Feb 20 '17 at 14:57
  • $\begingroup$ @SimplyBeautifulArt Thank you for you addition. Looking back, I probably should have provided a non-trivial example, so I am glad you have done so. $\endgroup$ – Chantry Cargill Feb 21 '17 at 14:49
  • $\begingroup$ Yeah, they aren't the easiest things to think of :D $\endgroup$ – Simply Beautiful Art Feb 21 '17 at 14:51

If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx$ exists, then we can use the following property

\begin{align} \int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx\qquad,\qquad\text{for }\, a>0. \end{align}


Using the property above for $a=1$, we can solve the following integral \begin{align} \int_{-\infty}^\infty \exp\left(-\frac{(x^2-qx-1)^2}{px^2}\right)\ dx &=\int_{-\infty}^\infty \exp\left(-\frac{(x-x^{-1}-q)^2}{p}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{(x-q)^2}{p}\right)\ dx\\ &=\int_{-\infty}^\infty \exp\left(-\frac{y^2}{p}\right)\ dy\\ &=\sqrt{p}\int_{-\infty}^\infty \exp\left(-z^2\right)\ dz\\ &=\sqrt{p\pi} \end{align} where $p>0$ and $q\in\mathbb{R}$.

Another technique, we may refer to Dirichlet integral, especially the double improper integral method.

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    $\begingroup$ More generally, $$\int_{-\infty}^\infty f(x)\,dx=\int_{-\infty}^\infty f\left(x-c-\sum_{i=1}^n \frac{a_i}{x-b_i}\right)\,dx$$ where $a_i>0$ for all $i$. $\endgroup$ – karvens Oct 11 '14 at 9:56
  • $\begingroup$ @karvens Wow! That's amazing. Thanks for sharing it here. BTW, how to prove it? $\endgroup$ – Anastasiya-Romanova 秀 Oct 11 '14 at 9:59
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    $\begingroup$ See orangeskid's answer here. $\endgroup$ – karvens Oct 11 '14 at 10:01
  • $\begingroup$ @karvens Ah, thanks! I forgot that answer although I also answered that OP, lol. $\endgroup$ – Anastasiya-Romanova 秀 Oct 11 '14 at 10:03
  • $\begingroup$ Is the last step wrong? Shouldn't it be $\sqrt{\frac{\pi}{p}}$ $\endgroup$ – Agile_Eagle Jul 19 '18 at 18:55

Here is a book for most advanced techniques

Advanced integration techniques

Here are some of the methods included in the book.

Laplace Integration

$$\int^{\infty}_0 \frac{f(t)}{t}\,dt=\, \int^{\infty}_0 \, \mathcal{L}(f (t))\, ds$$

Can be used to show that

$$\int^{\infty}_0 \frac{\sin(t)}{t}\,dt = \frac{\pi}{2}$$


$$\int^{\infty}_0 \frac{\sin(t)}{t}\,dt= \int^{\infty}_{0}\mathcal{L} (\sin(t))\,ds$$

We know that

$$\mathcal{L}(\sin(t)) = \frac{1}{s^2+1}$$

Just substitute in our integral

$$\int^{\infty}_0 \frac{ds}{1+s^2}= \tan^{-1}(s)|_{s=\infty}-\tan^{-1} (s)|_{s=0}=\frac{\pi}{2}$$ Convolution

$$\mathcal{L}\left((f * g)(t)\right)= \mathcal{L}(f(t)) \mathcal{L}(g (t)) $$

to show that

$$\beta(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma {(y+1)}}{\Gamma{(x+y+2)}}$$


Let us choose some functions $f(t) = t^{x} \,\, , \, g(t) = t^y$

Hence we get

$$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds $$

So by definition we have

$$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y ) $$

We can now use the laplace of the power

$$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$

Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$

$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{- 1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!} {(x+y+1)!}$$

So we have the following

$$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$

By definition we have

$$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds $$

This looks good , put $t=1$ we get

$$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$

By using that $n! = \Gamma{(n+1)}$

We arrive happily to our formula

$$ \int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma {(x+y+2)}}$$

which can be written as

$$ \int^{1}_0 s^{x-1}(1-s)^{y-1}\,ds= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma {(x+y+1)}}$$

Series expansion

$$\int^z_0 f(x) dx = \sum_{k\geq 0} a_k \frac{z^{k+1}}{k+1}$$

Can be used to show that

$$\int^\infty_0\frac{t^{s-1}}{e^t-1}dt = \Gamma(s) \zeta(s)$$


Using the power expansion

$$\frac{1}{1-e^{-t}} = \sum_{n=0}^\infty e^{-nt}$$

Hence we have

$$\int^\infty_0\,e^{-t}t^{s-1}\left(\sum_{n=0}^\infty e^{-nt}\right)\,dt$$

By swapping the series and integral

$$\sum_{n=0}^\infty\int^\infty_0\,t^{s-1}e^{-(n+1)t}\,dt = \Gamma(n) \sum_{n=0}^\infty \frac{1}{(n+1)^s}=\Gamma(s)\zeta(s)\,\,$$

Taking Limits

$$\lim_{s \to 0} \int^z_0 t^{s-1} f(t) \,dt = \int^z_0 \frac{f(t)}{t}dt$$

Can be used to prove

$$ \psi(a) = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz $$


Introduce the variable

$$\lim_{s \to 0} \int^{\infty}_0 (z^{s-1}e^{-z}-z^{s-1}(1+z)^{-a})\,dz $$

$$\Gamma(s)-\frac{\Gamma(s)\Gamma(a-s)}{\Gamma(a)} = \frac{\Gamma(s+1)}{\Gamma(a)}\left\{\frac{\Gamma(a)-\Gamma(a-s)}{s}\right\}$$

By taking the limit

$$\frac{1}{\Gamma(a)}\lim_{s \to 0}\frac{\Gamma(a)-\Gamma(a-s)}{s} =\frac{1}{\Gamma(a)}\lim_{s \to 0}\frac{\Gamma(a+s)-\Gamma(a)}{s} = \frac{\Gamma'(a)}{\Gamma(a)} = \psi(a)$$

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    $\begingroup$ Thanks I've been looking for a book like this for a long time. $\endgroup$ – Ahmed S. Attaalla Aug 8 '16 at 3:01
  • $\begingroup$ We usually don't like link-only answers, so can you put the major parts of your book into the answer? $\endgroup$ – Cyclohexanol. Aug 8 '16 at 3:13
  • $\begingroup$ @LaplacianFourier, I included some methods in the post from the book. $\endgroup$ – Zaid Alyafeai Aug 9 '16 at 20:51
  • $\begingroup$ To be honest I don't recall were I supposedly found a typo, and I retread your text multiple times but can't spot it. I'll let you know otherwise @ZaidAlyafeai $\endgroup$ – Ahmed S. Attaalla Sep 20 '16 at 23:35
  • $\begingroup$ @ZaidAlyafeai I'm only in high-school, that's why I didn't respond. It wouldn't be okay for someone with knowledge like me to be an editor. $\endgroup$ – Ahmed S. Attaalla Sep 24 '16 at 18:09

Odd/even function properties, symmetry of the function about a certain line.

$$\int^a_0f(x) dx=\int^a_0f(a-x) dx$$.

There are probably a couple of others that I have forgotten.

Edit: Never mind, I didn't notice that this asked for 'indefinite' integration- my apologies.

A neat trick is to sometimes multiply the integral by a factor of one (as is the case for integrating the secant and cosecant function).

  • $\begingroup$ I wish he hadn't specified indefinite to be honest. $\endgroup$ – Chantry Cargill Sep 23 '14 at 1:24
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    $\begingroup$ @Chantry Cargill, Yeah, you know what, indefinite is a little restricting isn't it? I'll edit the question to make it more general. $\endgroup$ – user3002473 Sep 23 '14 at 1:36

There are many integration techniques ranging from exact analytical methods like Contour Integration, change of variable, convolution techniques, stochastic integration... to approximate analytic methods using asymptotic expansions, continued franctions, Laplace's method ... but there's even more. A good detailed coverage of this material can be found in Daniel Zwillinger's The Handbook of Integration. If you find the latter to be too technical -- which may be the case if you didn't do any complex analysis -- then you may try George Boros / Victor Moll - Irresistible Integrals which would be a much friendlier read.


Differentiating under the integral. Albeit not technically "indefinite" integral technique, but you can still use it for those purposes.

Here is a link of how it is used:


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    $\begingroup$ The link is broken. $\endgroup$ – A---B Sep 8 '17 at 14:26

The use of residues to evaluate real integrals. There are many theorems that can be applied to various cases. If I get to it, I'll post some examples from this website. Later.


According to Wikipedia, the "tangent half-angle substitution" (a.k.a. the Weierstrauss substitution) is the "world's sneakiest substitution". It consists of subbing $x=\arctan(2t)$, which allows you to write $\sin(x) = 2t/(1+t^2)$, $\cos(x) = (1-t^2)/(1+t^2)$, and $dt = 2/(1+t^2) dx$. An example (poached from Wikipedia) is the following: $$ \begin{eqnarray} \int \csc(x) dx & = & \int \frac{dx}{\sin(x)} \\ & = & \int \left(\frac{2t}{1+t^2}\right)^{-1} \frac{2}{1+t^2} dt \\ & = & \int \frac{dt}{t} \\ & = & \ln(t) +C\\ & = & \ln(\tan(x/2)) + C \end{eqnarray} $$

Also, it's worth mentioning that for integrands which possess a symmetry, one can often evaluate the integral using ideas from representation theory. A simple example of this is the fact that the integral of an odd function over a domain symmetric about the origin vanishes. The symmetry here is parity.

A much more sophisticated example is when the integrand consists of functions with symmetry under a representation of the rotation group--e.g. spherical harmonics. The analysis of such integrals goes under the name "Wigner-Eckart Theorem", and is extremely important in physics.

The most fruitful application of this technique is usually to show that an integral is in fact $0$. In cases where an integral is non-zero, symmetry may tell you some information but not a complete answer. E.g. for an integral like $\int_{-1}^1 f(x) dx$ where $f(x)$ is even, you can't say what the exact value of the integral is, but you can say that it equals $2\int_0^1 f(x) dx$.

  • $\begingroup$ :O I can't believe the tangent half-angle substitution hasn't been mentioned yet! +1 $\endgroup$ – Simply Beautiful Art Jun 17 '17 at 14:33

I personally like the tactic of differentiation under the integral sign. Allow me to provide an example.

$$\frac{d}{du}\int_a^b f(u,x)\text{ dx}=\int_a^b \frac{\partial}{\partial u}f(u,x)\text{ dx}$$

This can be used to solve some extremely strange integrals, one of the easiest might be:

Prove $$I=\int_0^1 x\log x\text{ dx}=-\frac14$$

One solve this first by noticing for $a>1$

$$I(a):=\int_0^1 x^{a-1}\text{ dx}=a^{-1}.$$

We can note from this that $$I'(a)=\int_0^1 x^{a-1}\log x \text{ dx}=\left[a^{-1}\right]'=-a^{-2}$$

Evaluating $I'(a)$ at $a=2$ yields the upper result.


This help page lists the methods that the computer algebra system Maple uses: http://www.maplesoft.com/support/help/Maple/view.aspx?path=int/methods (full disclosure: I work for them). One interesting method that I, for one, didn't learn in university, is the Meijer G method. The idea here is that essentially all elementary and special functions can be expressed as so-called Meijer G functions, by using appropriate parameters. And there are relatively straightforward rules for finding the antiderivative of a Meijer G function. So that gives you your symbolic result in terms of Meijer G functions. The tricky part, then, is to express this in terms of the familiar standard functions and special functions.


There is one more useful integral relation from Fourier analysis; it's Plancherel's Theorem.

Let $f,g\in L^p$ and $\mathcal{F}_p$ the generalization of the Fouriertransformation $\mathcal{F}$ onto $L^p$ ($\mathcal{F}$ is usually only defined for $L^1$ or the Schwartz-space). Then, for any $p \ge 1$ Plancherel's identity

$\langle \mathcal{F}_p[f],\mathcal{F}_p[g]\rangle = \langle f, g\rangle$



It is

$$\mathcal{F}[1/(\lambda+i\kappa)](x) = \sqrt{2\pi}\Theta(x)e^{-\kappa x},\\ \mathcal{F}[e^{-\lambda^2/2}](x) = e^{-x^2/2}.$$

Therefore we find

$$\int_{-\infty}^\infty \frac{e^{-\lambda^2/2}}{\lambda + i \kappa} = \sqrt{2\pi} \int_0^\infty e^{-x^2/2 - \kappa x}.$$

The latter integral is readily solved by completing the square.

In general Plancherel's Theorem is usefull when computing the product of two seemingly unrelated functions (e.g. a rational and an exponential function), which have simplier representations in Fourierspace.


If we have $$u(x)=\prod_{i=1}^nf_i(x)$$ Then, through the product rule: $$u'(x)=\sum_{i=1}^{n}\prod_{j=1}^{n}f_j^{(\delta_{ij})}(x)$$ Where $\delta_{ij}$ is the Kronecker delta, $f^{(0)}(x)=f(x)$, and $f^{(1)}(x)=f'(x)$.

Thus, $$\int\sum_{i=1}^{n}\prod_{j=1}^{n}f_j^{(\delta_{ij})}(x)dx=\prod_{i=1}^nf_i(x)+C$$ It's really specific, and I haven't used it yet, but it's still a pretty neat formula (especially with that product notation).

Here's another. Assume $I=\{1,2,\dots,n\}$ for any $n\in\Bbb N$, and $\forall i,j\in I, i\neq j\iff \alpha_i\neq \alpha_j$ $$G=\int\prod_{i\in I}\frac{1}{x-\alpha_i}dx$$ Using partial fraction decomposition, we get $$G=\sum_{i\in I}\beta_{ij}\int\frac{dx}{x-\alpha_i}$$ Where $$\beta_{ij}=\prod_{i\neq j\in I}\frac{1}{\alpha_i-\alpha_j}$$ Then of course, $$G=\sum_{i\in I}\beta_{ij}\ln|x-\alpha_i|$$ Which is also pretty fun to look at (but is also little more useful).


protected by J. M. is a poor mathematician Mar 11 '18 at 14:26

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