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?
$\endgroup$ 17 Answers
$\begingroup$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.
$\endgroup$ 10 $\begingroup$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 ;)
$\endgroup$ 2 $\begingroup$you can try the most famous one which is:$$\int_0^\infty\frac{\sin(x)}{x}dx$$good luck!
$\endgroup$ 2 $\begingroup$Maybe you can look at:
Feynman's trick is used to compute:
\begin{align}\int_0^{\frac{\pi}{12}}\ln(\tan x)\,dx\end{align}
$\endgroup$ 2 $\begingroup$I know I am kind of late, but here are a few$$\int_{-\infty}^\infty\frac{\ln{(x^2+1)}}{x^4+x^2+1}\,dx$$$$\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}\,dx$$$$\int_{-\infty}^\infty\left(x^2+\frac{1}{x^2}\right)e^{-\left(x^2+\frac{1}{x^2}\right)}\,dx$$$$\int_{-\infty}^\infty\frac{x^2+2\cos{x}-2}{x^4}\,dx$$$$\int_{-\infty}^\infty\frac{x^2+2\cos{x}-2}{x^4(x^2+1)}\,dx$$$$\int_{-\infty}^\infty\left(\frac{1-\cos{x}}{x^2}\right)^2\,dx$$$$\int_0^\infty\frac{\ln{(x+\sqrt{x^2+1})}}{(x+\sqrt{x^2+1})^2}\,dx$$
$\endgroup$ $\begingroup$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$$\lim_{a\rightarrow1}I'\left(a\right)=\frac{\pi }{2e}.$$
$\endgroup$ $\begingroup$$I_9$ in Zacky's answer is a special case to$$I(s)=\int_0^\infty \frac{x^s}{x^2+1} \, {\rm d}x \tag{1}$$ with $-1<\Re(s)<1$, i.e.$$I_9 = \int_{2/3}^{4/5} I(s) \, {\rm d}s \, . $$(1) can be solved by contour integration$$2\pi i\, {\rm Res} \left( \frac{z^s}{z^2+1} \right)\Bigg|_{z=i}=\pi e^{i\pi s/2} = \oint_{-\infty}^\infty\frac{z^s}{z^2+1} \, {\rm d}z \\ = \int_{-\infty}^{-\epsilon}\frac{z^s}{z^2+1} \, {\rm d}z + \int_{|z|=\epsilon}\frac{z^s}{z^2+1} \, {\rm d}z + \int_\epsilon^\infty \frac{z^s}{z^2+1} \, {\rm d}z + \int_{|z|=\infty} \frac{z^s}{z^2+1} \, {\rm d}z \\ = I_- + I_\epsilon + I_+ + I_\infty \, ,$$where the contour is closed and avoids the cut in the upper half plane. The second and fourth integral can be estimated and $I_-$ can be related to $\lim_{\epsilon \rightarrow 0} I_+ = I(s)$$$|I_\epsilon| \leq \frac{\pi \epsilon^{s+1}}{1-\epsilon^2} \rightarrow 0 \quad \text{for} \quad \epsilon\rightarrow 0 \\ |I_\infty| \leq \frac{\pi R^{s+1}}{R^2-1} \rightarrow 0 \quad \text{for} \quad R\rightarrow \infty \\ \lim_{\epsilon \rightarrow 0} I_- \stackrel{z=-x}{=} \int_{0-i0}^{\infty-i0} \frac{(-x)^s}{x^2+1} \, {\rm d}x= e^{i\pi s} \int_0^\infty \frac{x^s}{x^2+1} \, {\rm d}x = e^{i\pi s} I(s) \, .$$Hence $$\pi e^{i\pi s/2} = \lim_{\epsilon \rightarrow 0} \left(I_+ + I_-\right) = (1+e^{i\pi s})I(s) \\ \Rightarrow \quad I(s)=\frac{\pi/2}{\cos(\pi s/2)} \, .$$Finally $$\int I(s) \, {\rm d}s = \log\left( \tan(\pi s/2) + \sec(\pi s/2) \right) + C \\ = \log\left( \tan\left(\frac{\pi}{4}(s+1)\right)\right) + C\, .$$
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