Tough integrals that can be easily beaten by using simple techniques

Question

This question is just idle curiosity. Today I find that an integral problem can be easily evaluated by using simple techniques like my answer to evaluate

\begin{equation} \int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx \end{equation}

I'm even shocked (and impressed, too) by user @Tunk-Fey's answer and user @David H's answer where they use simple techniques to beat hands down the following tough integrals

\begin{equation} \int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}\ dx \end{equation}

and

\begin{equation} \int_{0}^{\infty}\frac{\ln x}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\ dx \end{equation} So, I'm wondering about super tough definite integrals that can easily beaten by using simple techniques with only a few lines of answer. Can one provide it including its evaluation?

To avoid too many possible answers and to narrow the answer set, I'm interested in knowing tough integrals that can easily beaten by only using clever substitutions, simple algebraic manipulations, trigonometric identities, or the following property

\begin{equation} \int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx \end{equation}

I'd request to avoid using contour/ residue integrals, special functions (except gamma, beta, and Riemann zeta function), or complicated theorems. I'd also request to avoid standard integrals like

\begin{align} \int_{-1}^1\frac{\cos x}{1+e^{1/x}}\ dx&=\sin 1\tag1\\[10pt] \int_0^\infty\frac{\log ax}{b^2+c^2x^2}\ dx&=\frac{\pi\log\left(\!\frac{ab}{c}\!\right)}{2bc}\tag2\\[10pt] \int_0^1\frac{1}{(ax+b(1-x))^2}\ dx&=\frac{1}{bc}\tag3\\[10pt] \int_0^{\pi/2}\frac{\sin^kx}{\sin^kx+\cos^kx}\ dx&=\frac{\pi}{4}\tag4\\[10pt] \int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\ dx&=\log\left(\!\frac{b}{a}\!\right)\tag5 \end{align}

Answer 1

There is a lot of 'tough looking' integrals which can be solved by various tricks, but usually it requires more than a few lines of proof.

This is a really soft question, because 'tough looking' integral is a very subjective term (note that I use it instead of just 'tough' because I agree with Robert's comment).

I suggest the book Inside Interesting Integrals, it is full of tricks for solving definite integrals.

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My own example of solving an integral using only substitution and algebraic manipulations.

Solve (not really tough, but not simple either):

$$\int_0^{\infty} \frac{dx}{(1+x)(\pi^2+\ln^2 x)}$$

Let's introduce a parameter:

$$I(v)=\int_0^{\infty} \frac{dx}{(v+x)(\pi^2+\ln^2 x)}$$

Let's make a change of variable:

$$x=e^t$$

$$I(v)=\int_{-\infty}^{\infty} \frac{e^t dt}{(v+e^t)(\pi^2+t^2)}$$

$$I(v)=\int_{-\infty}^{\infty} \frac{(v+e^t) dt}{(v+e^t)(\pi^2+t^2)}-v \int_{-\infty}^{\infty} \frac{ dt}{(v+e^t)(\pi^2+t^2)}=1-v J(v)$$

Now let's make another change of variable:

$$t=-z$$

$$I(v)=\int_{-\infty}^{\infty} \frac{e^{-z} d(-z)}{(v+e^{-z})(\pi^2+z^2)}=\int_{-\infty}^{\infty} \frac{ dz}{(1+v e^z)(\pi^2+z^2)}=\frac{1}{v} J \left( \frac{1}{v} \right)$$

Now we get:

$$1-v J(v)=\frac{1}{v} J \left( \frac{1}{v} \right)=I(v)$$

$$v J(v)+\frac{1}{v} J \left( \frac{1}{v} \right)=1$$

We immediately get the correct value for the original integral:

$$J(1)=I(1)=\int_0^{\infty} \frac{dx}{(1+x)(\pi^2+\ln^2 x)}=\frac{1}{2}$$

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As another example, see this 2 line proof by Sangchul Lee of this:

$$\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0$$

This is problem C2.1 from the book I mentioned, but the book offers a very long and complicated solution. Which shows that simplicity of integrals is relative.

Answer 2

My favourite example of this is @SangchulLee's solution to @VladimirReshetnikov's question, which asks to verify the correctness of the identity $$\int_0^{\infty} \frac{dx}{\sqrt[4]{7 + \cosh x}}= \frac{\sqrt[4]{6}}{3\sqrt{\pi}} \Gamma\left(\frac14\right)^2 .$$

The other answers indicate the "toughness" of this integral, resorting to all sorts of special functions such as elliptic functions, hypergeometric functions, or Mathematica.

However, the integral can be brilliantly shown to be a few substitutions away from the form of a beta function integral.

Lee makes a chain of simple substitutions which he desecribes here, to obtain $$\int_0^{\infty} \frac{dx}{(a+ \cosh x)^s} = \frac1{(a+1)^s} \int_0^1 \frac{v^{s-1}}{\sqrt{(1-v)(1-\frac{a-1}{a+1} v)}} dv.$$

The closed formed of this integral for general $a$ is certainly non-elementary, but our special case $a=7$ and $s=4$ is different for a very neat reason:

When $a=7,$ $\frac{a-1}{a+1}$ is equal to $\frac34$, but since we have the triple angle formula $\displaystyle \, \cosh(3 x)=4\cosh^3 x-3 \cosh x,$ the integral can be rewritten (with $v=\operatorname{sech}^2 t$) as $$2^{5/4} \int_0^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} dt$$ which can be easily brought to the form of a beta function.

(Note that we can find a similar closed form (with $a=7$) for $s=3/4$.)

Answer 3

Ramanujan's master theorem can be applied to a wide range of (sometimes extremely complicated) definite integrals, allowing them to be evaluated in less than a line of computations. The theorem states:

If $f(x)$ has a series expansion of the form:

$$f(x) = \sum_{k=0}^{\infty}(-1)^{k}\frac{\lambda(k)}{k!} x^k$$

then

$$\int_0^{\infty}x^{s-1}f(x) dx = \Gamma(s)\lambda(-s)$$

Here one has to use an appropriate analytic continuation of $\lambda(k)$. This is a generalization of an old theorem obtained by Glaisher stating that:

$$\int_0^{\infty}\sum_{k=0}^{\infty}(-1)^k a_k x^{2k}dx = \frac{\pi}{2}a_{-\frac{1}{2}}$$

where again an appropriate analytic continuation of the expansion coefficients is assumed.

Example: From the series expansion of the Bessel function of the first kind $J_{\alpha}(x)$ it is very easy to compute that:

$$\int_0^{\infty}x^p J_{\alpha}(x)dx = 2^p\frac{\Gamma\left(\frac{\alpha + 1 +p}{2}\right)}{\Gamma\left(\frac{\alpha + 1 -p}{2}\right)}$$

for $-(\alpha + 1)<p<\frac{1}{2}$

Answer 4

Feynman's trick, Frullani's theorem, symmetry tricks, Glasser's master theorem, the Laplace transform, Fourier (Legendre/Chebyshev) series expansions and the Euler beta function gave me at least the $80\%$ of my reputation points, but the day still has to come, to be prouder than this massacre through the residue theorem.

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I have just realized I forgot to mention Wilf's wonder: creative telescoping.

Answer 5

Some integrals I would consider:

1. $\int(\frac{x^4}{1+ x^6})^2 dx$ . This integral involves a very interesting trigonometric substitution.

2. $\int[\ln(x)\arcsin(x)] dx$. It is one of the few integrals which contain a logarithmic and an inverse trig function. It is an interesting one.

3. $\int \sin(x)\arctan(\sqrt{\sec(x)-1}) dx$. Once again, another interesting trigonometric substitution.