# integrals

#### sbhatnagar

##### Active member
integrals (lots of them!)

1. $\displaystyle \int_{0}^{1}\frac{\sin(\ln x)}{\ln x}dx$

2. $\displaystyle \int_{0}^{1}\frac{\arctan(x)}{1+x}dx$

3. $\displaystyle \int_{0}^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx$

4. $\displaystyle \int_{0}^{\infty}e^{- \left( x^2+\dfrac{1}{x^2}\right)}dx$

5. $\displaystyle \int_{0}^{2\pi}e^{\cos \theta} \cos(\sin \theta) d\theta$

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#### chisigma

##### Well-known member
Re: integrals (lots of them!)

Remembering that is...

$\displaystyle \mathcal{L}\{\frac{sin t}{t}\}= \int_{s}^{\infty} \frac{du}{1+u^{2}}= \cot^{-1} s$ (1)

... we obtain...

$\displaystyle \int_{0}^{1} \frac{\sin \ln x}{\ln x}\ dx= \int_{-\infty}^{0}\frac{\sin t}{t}\ e^{t}\ dt = \int_{0}^{\infty}\frac{\sin t}{t}\ e^{-t}\ dt = \cot^{-1} 1= \frac{\pi}{4}$ (2)

Kind regards

$\chi$ $\sigma$

#### Sherlock

##### Member
Re: integrals (lots of them!)

Keep the good stuff coming!

2. $\displaystyle \int_{0}^{1}\frac{\arctan(x)}{1+x}dx$
Sub $x = \tan{t}$, we get

\displaystyle \begin{aligned}I & = \int_{0}^{1}\frac{ \arctan{x}}{1+x}\;{dx} \\& = \int_{0}^{\pi/4}\frac{t \sec{t}}{\cos{t}+\sin{t}}\;{dt}.\end{aligned}

Put $t \mapsto \frac{\pi}{4}-t$ and add them to get:

\displaystyle \begin{aligned} 2I & = \frac{\pi}{4}\int_{0}^{\pi/4}\frac{\sec{t}}{\cos{t}+\sin{t}}\;{dt} \\& = \frac{\pi}{4}\int_{0}^{\pi/4}\frac{\sec^2{t}}{1+\tan{t}}\;{dt} \\& = \frac{\pi}{4}\int_{0}^{\pi/4}\frac{(1+\tan{t})'}{1+\tan{t}}\;{dt} \\& = \frac{\pi}{4}\ln\bigg|1+\tan{t}\bigg|_{0}^{\pi/4} \\& = \frac{\pi}{4}\ln(2). \end{aligned}

Therefore $\displaystyle \int_{0}^{1}\frac{\arctan(x)}{1+x}\;{dx} = \frac{\pi}{8}\ln(2).$

3. $\displaystyle \int_{0}^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx$
Sub $x = \tan(\varphi)$ then put $\varphi \mapsto \frac{\pi}{2}-\varphi$, we get

\displaystyle \begin{aligned} I & = \int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2}\;{dx} \\& = -2\int_{0}^{\pi/2}\ln(\cos\varphi)\;{d\varphi} \\& = -2\int_{0}^{\pi/2}\ln(\sin\varphi)\;{d\varphi}. \end{aligned}

Adding the second two integrals we get

\displaystyle \begin{aligned}2I & = -2\int_{0}^{\pi/2}\ln\left(\frac{1}{2}\sin 2 \varphi\right)\;{d\varphi} \\& =\pi\ln(2)-2\int_{0}^{\pi/2}\ln(\sin 2\varphi)\;{d\varphi} \\& = \pi\ln(2)-\int_{0}^{\pi}\ln(\sin\varphi)\;{d\varphi} \\& = \pi\ln(2)-2\int_{0}^{\pi/2}\ln(\sin\varphi)\;{d\varphi} \\& = \pi\ln(2)+I.\end{aligned}

Therefore $\displaystyle \int_{0}^{\infty} \frac{\ln(1+x^2)}{1+x^2}\;{dx} = \pi\ln(2).$

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#### sbhatnagar

##### Active member
Re: integrals (lots of them!)

$\displaystyle \int_{0}^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx$
Here is my approach for the third one.

$\displaystyle I(\lambda)=\int_{0}^{\infty}\frac{\ln(1+\lambda x^2)}{1+x^2}dx$

$\displaystyle I'(\lambda)=\int_{0}^{\infty}\frac{x^2}{(1+\lambda x^2)(1+x^2)}dx=\frac{1}{1-\lambda}\int_{0}^{\infty}\frac{1}{1+\lambda x^2}dx+\frac{1}{\lambda-1}\int_{0}^{\infty}\frac{1}{1+x^2}dx$

$\displaystyle =\frac{\pi}{2(\lambda-1)}-\frac{\pi}{2\sqrt{\lambda}(\lambda-1)}=\frac{\pi}{2(\lambda+\sqrt{\lambda})}$

Integrating both side yields:

$\displaystyle I(\lambda)=\pi \ln(\sqrt{\lambda}+1)+C$

For $\lambda=0$ we have $C=0$.

$\displaystyle I(\lambda)=\pi \ln(\sqrt{\lambda}+1)$

Therefore

$\displaystyle \int_{0}^{\infty}\frac{\ln(1+x^2)}{1+x^2}dx=I(1)= \pi \ln(2)$

---------- Post added at 02:34 PM ---------- Previous post was at 02:11 PM ----------

$\displaystyle \int_{0}^{1}\frac{\sin( \ln x)}{\ln x}dx$
For the first one I did the following:

$\displaystyle I(\lambda)=\int_{0}^{1}\frac{\sin(\lambda \ln x)}{\ln x}dx$

$\displaystyle I'(\lambda)=\int_{0}^{1}\cos(\lambda \ln x)dx=\int_{-\infty}^{0}e^x \cos(\lambda x)dx$

$\displaystyle =\left[ \frac{e^x \{ \cos(\lambda x)+\lambda \sin(\lambda x)\}}{1+\lambda^2}\right]_{-\infty}^{0}=\frac{1}{1+\lambda^2}$

Integrate both sides:

$\displaystyle I(\lambda)=\arctan(\lambda)+C$

For $\lambda=0$ we have $C=0$. so

$\displaystyle I(\lambda)=\arctan(\lambda)$

Therefore

$\displaystyle \int_{0}^{1}\frac{\sin( \ln x)}{\ln x}dx=I(1)=\arctan(1)=\frac{\pi}{4}$

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#### chisigma

##### Well-known member
Re: integrals (lots of them!)

Let's define...

$\displaystyle \varphi(t)=\int_{0}^{\infty} e^{-(x^{2}+\frac{t^{2}}{x^{2}})}\ dx$ (1)

... and now we derive (1)...

$\displaystyle \varphi^{'}(t)=-2\ t\ \int_{0}^{\infty} \frac{1}{x^{2}}\ e^{-(x^{2}+\frac{t^{2}}{x^{2}})}\ dx$ (2)

Setting in (2) $\displaystyle \tau=\frac{t}{x}$ in some steps we obtain...

$\displaystyle \varphi^{'}(t)=-2\ \int_{0}^{\infty} e^{-(\tau^{2}+\frac{t^{2}}{\tau^{2}})}\ d \tau = -2\ \varphi(t)$ (3)

The (3) is an easily solvable ODE and, taking into account the well known result...

$\displaystyle \int_{0}^{\infty} e^{-x^{2}}\ dx = \frac{\sqrt{\pi}}{2}$ (4)

... we obtain...

$\displaystyle \varphi(t)= \frac{\sqrt{\pi}}{2}\ e^{-2\ t}$ (5)

Setting in (5) t=1 we finally obtain...

$\displaystyle \int_{0}^{\infty} e^{-(x^{2}+\frac{1}{x^{2}})}\ dx =\frac{\sqrt{\pi}}{2}\ e^{-2}$ (6)

Kind regards

$\chi$ $\sigma$

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#### sbhatnagar

##### Active member
Re: integrals (lots of them!)

$\displaystyle \int_{0}^{\infty}e^{-\left(\displaystyle x^2+\dfrac{1}{x^2} \right)}dx$
Yeah that's right. I also used the same technique. Here is another solution which amazed me:

$\displaystyle I=\int_{0}^{\infty}e^{-\left(\displaystyle x^2+\dfrac{1}{x^2} \right)}dx=\int_{0}^{\infty}e^{-\left(\displaystyle x^2+\dfrac{1}{x^2}-2 \right)-\displaystyle 2}dx=e^{-2}\int_{0}^{\infty}e^{-\left(\displaystyle x^2+\dfrac{1}{x^2}-2 \right)}dx$

Use the fact that $\displaystyle \int_{0}^{\infty}f \left(x^2+\frac{1}{x^2}-2\right)dx=\int_{0}^{\infty}f \left(x^2\right)dx$:

$\displaystyle I=e^{-2}\int_{0}^{\infty}e^{-x^2}dx=e^{-2}\frac{\sqrt{\pi}}{2}$

#### Sherlock

##### Member
Re: integrals (lots of them!)

You guys took the term 'magic differentiation' to a whole new level!

#### sbhatnagar

##### Active member
Re: integrals (lots of them!)

Solution for 5.

$$\int_{0}^{2\pi}e^{\cos \theta}\cos(\sin \theta)d\theta$$
$\displaystyle I(\phi)=\int_{0}^{2\pi}e^{\phi \cos \theta}\cos(\phi \sin \theta)d\theta$

$\displaystyle I'(\phi)=\int_{0}^{2\pi}e^{\phi \cos \theta}(\cos(\theta)\cos(\phi \sin \theta)-\sin(\theta)\sin(\phi \sin \theta))d\theta$

$\displaystyle =\frac{1}{\phi}\int_{0}^{2\pi}\frac{d}{d\theta}(e^{\phi\cos \theta} \sin(\phi \sin\theta))d\theta=\frac{1}{\phi}\int_{0}^{2\pi}d(e^{\phi\cos \theta} \sin(\phi \sin\theta))$

$\displaystyle =0$

Integrate both sides:

$\displaystyle I(\phi)=0+C$

When $\phi=0$ then $C=2\pi$.

Therefore

$\displaystyle I(\phi)=2\pi$

so

$\displaystyle \int_{0}^{2\pi}e^{\cos \theta}\cos(\sin \theta)d\theta =I(1)=2\pi$