Integration using the Feynman Trick

I have a feeling this post won't met the community guidelines (will delete if so).

I'm looking for definite integrals that are solvable using the method of differentiation under the integral sign (also called the Feynman Trick) in order to practice using this technique.

Does anyone know of any good ones to tackle?

• @Q the Platypus - thanks for the edit. – user150203 Nov 7 '18 at 1:56

Here are some that I have encountered: $$I_1=\int_0^\frac{\pi}{2} \ln(\sec^2x +\tan^4x)dx$$ $$I_2=\int_0^\infty \frac{\ln\left({1+x+x^2}\right)}{1+x^2}dx$$ $$I_3=\int_0^\frac{\pi}{2}\ln(2+\tan^2x)dx$$ $$I_4=\int_0^\infty \frac{x-\sin x}{x^3(x^2+4)} dx$$ $$I_5=\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ $$I_6=\int_0^\frac{\pi}{2} \ln\left(\frac{2+\sin x}{2-\sin x}\right)dx$$ $$I_7=\int_0^\frac{\pi}{2} \frac{\arctan(\sin x)}{\sin x}dx$$ $$I_8=\int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx$$ $$I_9=\int_0^{\infty} \frac{x^{4/5}-x^{2/3}}{\ln(x)(1+x^2)}dx$$ $$I_{10}=\int_0^1 \frac{\ln(1+x)}{x(1+x^2)}dx$$ $$I_{11}=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\,, a=2; a=\frac12$$ $$I_{12}=\int_0^1 \frac{\ln(1-x+x^2)}{x(1-x)}dx$$

In case you struggle where to put that parameter, feel free to ask.

• I am working my way through the list (thanks again). I solved $I_4$ yesterday math.stackexchange.com/questions/3004469/… – user150203 Nov 20 '18 at 4:16
• That is nice to hear! I will try to add more when I have time. – Zacky Nov 20 '18 at 9:59
• Have been coming back and forth with these. Got $I_7$ out yesterday. Was very close to $I_1$ too! math.stackexchange.com/questions/3024869/… – user150203 Dec 3 '18 at 23:27
• That is nice to hear, as for $\,I_1=\int_0^\frac{\pi}{2} \ln(\sec^2x +\tan^4x)dx\,$ I tried to keep it in the original way, but after the substitution $\tan x =t$ might be clearer how to use Feynman's trick. – Zacky Dec 4 '18 at 0:40
• I finally got $I_1$ out! math.stackexchange.com/questions/3026362/… – user150203 Dec 4 '18 at 23:46

A few good ones are: $$\int_0^\infty e^{-\frac{x^2}{y^2}-y^2}dx$$ $$\int_0^\infty \frac{1-\cos(xy)}xdx$$ $$\int_0^\infty \frac{dx}{(x^2+p)^{n+1}}$$ $$\int_{0}^{\infty}e^{-x^2}dx$$ $$\int_0^\infty \cos(x^2)dx$$ $$\int_0^\infty \sin(x^2)dx$$ $$\int_0^\infty \frac{\sin^2x}{x^2(x^2+1)}dx$$ $$\int_0^{\pi/2} x\cot x\ dx$$ That should keep you busy for a while ;)

Maybe you can look at:

https://math.stackexchange.com/a/2989801/186817

Feynman's trick is used to compute:

\begin{align}\int_0^{\frac{\pi}{12}}\ln(\tan x)\,dx\end{align}

• Is the upper limit meant to be $\frac{\pi}{2}$ – user150203 Nov 16 '18 at 1:00
• No, it's $\dfrac{\pi}{12}$. with upper bound to be $\dfrac{\pi}{2}$: \begin{align}\int_0^{\frac{\pi}{2}}\ln(\tan x)\,dx=0\end{align} (perform the change of variable $y=\dfrac{\pi}{2}-x$ ) – FDP Nov 16 '18 at 15:28

you can try the most famous one which is: $$\int_0^\infty\frac{\sin(x)}{x}dx$$ good luck!

Another example is in evaluating $$\displaystyle \int_0^\infty \dfrac{\cos xdx}{1+x^2}$$

by first considering $$I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(ax\right)}{x\left(1+x^{2}\right)}dx,\,a>0$$ we have $$I'\left(a\right)=\int_{0}^{\infty}\frac{\cos\left(ax\right)}{1+x^{2}}dx$$ From which it can be shown that $$I\left(a\right)=\frac{\pi}{2}\left(1-e^{-a}\right)$$ hence $$\lim_{a\rightarrow1}I'\left(a\right)=\frac{\pi }{2e}.$$