Definite integrals solvable 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?

Solution 1:

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.

Solution 2:

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 ;)

Solution 3:

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

Solution 4:

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}

Solution 5:

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 $$\lim_{a\rightarrow1}I'\left(a\right)=\frac{\pi }{2e}.$$