Using differentiation under integral sign to calculate $\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$

Question

I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$$

I know that I have to solve from differentiate under the integral and I must use a suitable substitution for integrands involving trigonometric functions but I can't complete the solution. Could you help me?

Answer 1

Define

$$I(a) = \int_0^{\pi/2} d\phi \frac{\log{(1+a\sin{\phi})}}{\sin{\phi}} $$

Then

$$I'(a) = \int_0^{\pi/2} \frac{d\phi}{1+a\sin{\phi}} $$

To evaluate the latter integral, sub $t=\tan{\phi/2}$. Then $d\phi = 2/(1+t^2) dt$ (why?), and $\sin{\phi} = 2 t/(1+t^2)$ (why again?), and the integral is

$$I'(a) = 2 \int_0^1 \frac{dt}{1+t^2} \frac1{1+2 a t/(1+t^2)} = 2 \int_0^1 \frac{dt}{1+2 a t+t^2}$$

This is easily evaluated by completing the square in the denominator:

$$\begin{align}I'(a) &= 2 \int_0^1 \frac{dt}{(t+a)^2+1-a^2} \\ &= \frac{2}{\sqrt{1-a^2}} \left [\arctan{\frac{t+a}{\sqrt{1-a^2}}} \right ]_0^1 \\ &=\frac{2}{\sqrt{1-a^2}} \left [\arctan{\frac{1+a}{\sqrt{1-a^2}}}- \arctan{\frac{a}{\sqrt{1-a^2}}}\right ]\\ &= \frac{2}{\sqrt{1-a^2}} \arctan{}\frac{\sqrt{1-a^2}}{1+a+a^2} \end{align}$$

Now we must integrate with respect to $a$. To do this, we integrate over each of the separate terms in the penultimate line. To begin

$$\int da \, \frac{2}{\sqrt{1-a^2}} \arctan{\frac{a}{\sqrt{1-a^2}}}$$

Sub $a=\sin{\theta}$, and it should be straightforward to see that this antiderivative is simply

$$2 \int d\theta \, \theta = \theta^2 = \arcsin^2{a} $$

(Yes I am ignoring the constant of integration for now.)

Now the second piece:

$$\int da \, \frac{2}{\sqrt{1-a^2}} \arctan{\frac{1+a}{\sqrt{1-a^2}}}$$

Sub $a=\cos{\theta} $. Note that the argument of the arctan becomes $\cot{(\theta/2)}$, so the integral becomes

$$-2 \int d\theta \left (\frac{\pi}{2} - \frac{\theta}{2} \right ) = \frac12 \theta^2 - \pi \theta = \frac{\pi}{2} \arcsin{a} + \frac12 \arcsin^2{a}-\frac{3 \pi^2}{8}$$

Including the constant of integration, then,

$$I(a) = \frac12 \theta^2 - \pi \theta = \frac{\pi}{2} \arcsin{a} - \frac12 \arcsin^2{a}-\frac{3 \pi^2}{8} + C$$

Given $I(0)=0$, then $I(a) = \frac{\pi}{2} \arcsin{a} - \frac12 \arcsin^2{a}$. The desired integral is $I(1)$, so that

$$\int_0^{\pi/2} d\phi \frac{\log{(1+\sin{\phi})}}{\sin{\phi}} = \frac{\pi^2}{4} - \frac{\pi^2}{8} = \frac{\pi^2}{8} $$

EDIT

I should have mentioned above that $a$ is restricted to values such that $|a| \le 1$ in the above analysis.

Answer 2

Let's consider the integral

\begin{align}I(\alpha)&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\sin\,\phi)}{\sin\,\phi}\;d\phi\quad\Rightarrow\quad\phi\mapsto \frac{\pi}{2}-\phi\\ &=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\cos\,\phi)}{\cos\,\phi}\;d\phi, \qquad 0 < \alpha < \pi.\end{align}

Differentiating $I(\alpha)$ with respect to $\alpha$, we have

\begin{align} {I}'(\alpha) &= \int_0^{\Large\frac{\pi}{2}} \frac{\partial}{\partial\alpha} \left(\frac{\ln(1 + \cos\alpha \cos \phi)}{\cos \phi}\right)\,d\phi \\ &=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{1+\cos \alpha \cos \phi}\,d\phi \\ &=-\int_0^{\Large\frac{\pi}{2}}\frac{\sin \alpha}{\left(\cos^2 \frac{\phi}{2}+\sin^2 \frac{\phi}{2}\right)+\cos \alpha\,\left(\cos^2\,\frac{\phi}{2}-\sin^2 \frac{\phi}{2}\right)}\,d\phi \\ &=-\frac{\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}} \frac{1}{\cos^2\frac{\phi}{2}}\frac{1}{\left[\left(\frac{1+\cos \alpha}{1-\cos \alpha}\right) +\tan^2 \frac{\phi}{2} \right]}\,d\phi \\ &=-\frac{2\,\sin\alpha}{1-\cos\alpha} \int_0^{\Large\frac{\pi}{2}}\,\frac{\frac{1}{2}\,\sec^2\,\frac{\phi}{2}}{\left[\,\left(\dfrac{2\,\cos^2\,\frac{\alpha}{2}}{2\,\sin^2\,\frac{\alpha}{2}}\right) + \tan^2\,\frac{\phi}{2} \right]} \,d\phi \\ &=-\frac{2\left(2\,\sin\,\frac{\alpha}{2}\,\cos\,\frac{\alpha}{2}\right)}{2\,\sin^2\,\frac{\alpha}{2}}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\left(\dfrac{\cos \frac{\alpha}{2}}{\sin\,\frac{\alpha}{2}}\right)^2\,+\,\tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan\,\frac{\phi}{2}\right)\\ &=-2\cot \frac{\alpha}{2}\,\int_0^{\Large\frac{\pi}{2}}\,\frac{1}{\left[\,\cot^2\,\frac{\alpha}{2} + \tan^2\,\frac{\phi}{2}\,\right]}\,d\left(\tan \frac{\phi}{2}\right)\,\\ &=-2\,\left.\tan^{-1} \left(\tan \frac{\alpha}{2} \tan \frac{\phi}{2} \right) \right|_0^{\Large\frac{\pi}{2}}\\ &=-\alpha \end{align}

Therefore:

$$I(\alpha) = C - \frac{\alpha^2}{2}$$

However by definition, $I\left(\frac{\pi}{2}\right) = 0$, hence $C = \dfrac{\pi^2}{8}$ and

$$I(\alpha) = \frac{\pi^2}{8}-\frac{\alpha^2}{2}.$$

The integral we want to evaluate is

$$I(0) = \int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\sin\,\phi)}{\sin\,\phi}\;d\phi=\frac{\pi^2}{8}.$$

Answer 3

We can use the following Taylor expansion \begin{eqnarray} \ln(1+x)&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}x^n. \end{eqnarray} Then \begin{eqnarray} \int_0^{\pi/2}\frac{\ln(1+\sin\phi)}{\sin\phi}d\phi&=&\int_0^{\pi/2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\sin^{n-1}\phi d\phi\\ &=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\frac{\sqrt{\pi}\Gamma(\frac{n}{2})}{2\Gamma(\frac{n+1}{2})}\\ &=&\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{(2n-1)\Gamma(n)}-\frac{\sqrt{\pi}}{4}\sum_{n=1}^\infty\frac{\Gamma(n)}{n\Gamma(\frac{2n+1}{2})}\\ \end{eqnarray} Let $$ f(x)=\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{(2n-1)\Gamma(n)}x^{2n}, g(x)=\frac{\sqrt{\pi}}{4}\sum_{n=1}^\infty\frac{\Gamma(n)}{n\Gamma(\frac{2n+1}{2})}x^{2n}. $$ Then $$ \left(\frac{f(x)}{x}\right)'=\frac{\sqrt{\pi}}{2}\sum_{n=1}^\infty\frac{\Gamma(\frac{2n-1}{2})}{\Gamma(n)}x^{2n-2}=\frac{\pi}{2\sqrt{1-x^2}} $$ and hence $$ f(x)=\frac{\pi}{2} x\arcsin x. $$ Also $$ g(x)=\frac{1}{2}\arcsin^2x. $$ So $f(1)=\frac{\pi^2}{4}, g(1)=\frac{\pi^2}{8}$ and hence $$ \int_0^{\pi/2}\frac{\ln(1+\sin\phi)}{\sin\phi}d\phi=\frac{\pi^2}{8}. $$