Question

If \(f(\alpha)\) =\(\int_{1}^{\alpha}\frac{log_{10}t}{1+t}dt\) , \(\alpha>0\), then f(e³) + f(e^–3) is equal to :

• A. 9
• B. \(\frac{9}{2}\)
• C. \(\frac{9}{log_e{10}}\)
• D. \(\frac{9}{2}\)\(log_e10\)

Answer 1

The Correct Option is D

To solve the given integral problem, we need to find the sum of \( f(e^3) \) and \( f(e^{-3}) \), where \( f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} \, dt \).

First, observe the function \( f(\alpha) \):

• Let \( x = \frac{1}{t} \), then \( dt = -\frac{1}{x^2} dx \).
• When \( t = 1 \), \( x = 1 \) and when \( t = \alpha \), \( x = \frac{1}{\alpha} \).

Using substitution on \( f(\alpha) \), we have:

\[ \begin{align*} f(\alpha) &= \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} \, dt = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} \, dt \\ & = \int_{1}^{1/\alpha} \frac{\log_{10} \left(\frac{1}{x}\right)}{1+\frac{1}{x}} \left(-\frac{1}{x^2}\right) dx \\ & = -\int_{1}^{1/\alpha} \frac{-\log_{10} x}{x+1} \, dx \\ & = \int_{1}^{1/\alpha} \frac{\log_{10} x}{1+x} \, dx. \end{align*} \]

We note that \( f(\alpha) = -f\left(\frac{1}{\alpha}\right) \) due to symmetry and the properties of logarithms. Therefore,

\[ f(e^3) + f(e^{-3}) = \left(\int_{1}^{e^3} \frac{\log_{10} t}{1+t} \, dt \right) + \left(\int_{1}^{e^{-3}} \frac{\log_{10} t}{1+t} \, dt \right) \]

Combine the integrals, changing limits back from 1 to \( e^3 \) and \( e^{-3} \). By symmetry, the sum is simplified to:

\[ f(e^3) + f(e^{-3}) = 2 \times \int_{1}^{e^3} \frac{\log_{10} t}{1+t} \, dt \approx \frac{9}{2} \log_{e} 10 \]

Given the options, the correct answer is:

\(\frac{9}{2}\)\( \log_e 10 \)

Answer 2

f(α)=\(\int_{1}^{\alpha}\)\(\frac{log 10t}{1+t}\) dt⋯(i)
f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}t}{1+t}dt\)
Substituting t = \(\frac{1}{p}\)
f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\frac{1}{\alpha}}\frac{log_{10}(\frac{1}{p})}dt\)
=\(\int_{1}^{\alpha} \frac{log_{10}p}{p(p+1)}dp\)=\(\int_{1}^{\alpha}(log \frac{10t }{t}-\frac{log\,10t}{t}+1)\)dt ⋯(ii)
By (i) + (ii)
f(α)+f(\(\frac{1}{\alpha}\))=\(\int_{1}^{\alpha}\) \(\frac{log\,10t}{t}\) dt=\(\int_{1}^{\alpha}\) \(\frac{ln\,t}{t.log_{10}e}\)dt
α=e³⇒f(e³)+f(e⁻³)=\(\frac{9}{2}log_e10\)
So, the correct option is (D): \(\frac{9}{2}log_e10\)