{"id":20130,"date":"2024-04-15T05:48:47","date_gmt":"2024-04-15T05:48:47","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20130"},"modified":"2024-04-15T05:48:47","modified_gmt":"2024-04-15T05:48:47","slug":"the-value-of-int_0infty-frac11-textx2-textdx-int_0infty-fracsin-textxtextx-textdx-is-a-fracpi-2-b-pi-c","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-value-of-int_0infty-frac11-textx2-textdx-int_0infty-fracsin-textxtextx-textdx-is-a-fracpi-2-b-pi-c\/","title":{"rendered":"The value of $$\\int_0^\\infty {\\frac{1}{{1 + {{\\text{x}}^2}}}} {\\text{dx}} + \\int_0^\\infty {\\frac{{\\sin {\\text{x}}}}{{\\text{x}}}} {\\text{dx}}$$ is A. $$\\frac{\\pi }{2}$$ B. $$\\pi $$ C. $$\\frac{{3\\pi }}{2}$$ D. 1"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a9890d12cfb4\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"true\">\r\n                    $$\frac{pi }{2}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    $$pi $$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    $$\frac{{3pi }}{2}$$                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"false\">\r\n                    1                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The value of $$\\int_0^\\infty {\\frac{1}{{1 + {{\\text{x}}^2}}}} {\\text{dx}} + \\int_0^\\infty {\\frac{{\\sin {\\text{x}}}}{{\\text{x}}}} {\\text{dx}}$$ is $\\pi$.<\/p>\n<p>To see this, we can use the residue theorem. Consider the following contour integral:<\/p>\n<p>$$\\oint_C dz \\frac{1}{z(1+z^2)}$$<\/p>\n<p>where $C$ is a keyhole contour in the complex plane, as shown below.<\/p>\n<p>[asy]<br \/>\nunitsize(1 cm);<\/p>\n<p>draw((0,-1.2)&#8211;(0,1.2));<br \/>\ndraw((-1.2,0)&#8211;(1.2,0));<\/p>\n<p>real g(real x) {<br \/>\n  return sqrt(1+x^2);<br \/>\n}<\/p>\n<p>draw(graph(g,-1.2,1.2),red);<\/p>\n<p>draw((0,0)&#8211;(1.2,0.6));<br \/>\ndraw((0,0)&#8211;(-1.2,0.6));<\/p>\n<p>draw((1.2,0.6)&#8211;(1.2,-0.6),dashed);<br \/>\ndraw((-1.2,0.6)&#8211;(-1.2,-0.6),dashed);<\/p>\n<p>label(&#8220;$Re(z)$&#8221;, (1.2,0.2), E);<br \/>\nlabel(&#8220;$Im(z)$&#8221;, (0.2,1.2), N);<\/p>\n<p>label(&#8220;$z=0$&#8221;, (0,0), S);<br \/>\nlabel(&#8220;$z=i$&#8221;, (0,1), E);<br \/>\nlabel(&#8220;$z=-i$&#8221;, (0,-1), W);<br \/>\n[\/asy]<\/p>\n<p>The contour integral is equal to the sum of the integrals over the four line segments. The integrals over the vertical line segments go to zero as the radius of the semicircles goes to infinity. Therefore, the contour integral is equal to<\/p>\n<p>$$\\int_0^\\infty dx \\frac{1}{1+x^2} + i \\int_0^\\infty dx \\frac{x}{1+x^2} + \\int_{-\\infty}^0 dx \\frac{1}{1+x^2} &#8211; i \\int_{-\\infty}^0 dx \\frac{x}{1+x^2}$$<\/p>\n<p>The real part of the contour integral is <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div> therefore<\/p>\n<p>$$\\int_0^\\infty dx \\frac{1}{1+x^2} &#8211; \\int_{-\\infty}^0 dx \\frac{1}{1+x^2}$$<\/p>\n<p>Using the substitution $x \\mapsto -x$, we can write this as<\/p>\n<p>$$2 \\int_0^\\infty dx \\frac{1}{1+x^2}$$<\/p>\n<p>The imaginary part of the contour integral is zero.<\/p>\n<p>Therefore, the residue theorem tells us that<\/p>\n<p>$$2 \\pi i \\cdot 2 \\pi i \\cdot 0 = \\pi \\cdot \\frac{1}{2} + \\pi \\cdot \\frac{-1}{2}$$<\/p>\n<p>or<\/p>\n<p>$$\\int_0^\\infty dx <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> \\frac{1}{1+x^2} = \\pi$$<\/p>\n<p>Now, we can use the substitution $x \\mapsto \\sin \\theta$ to write<\/p>\n<p>$$\\int_0^\\infty dx \\frac{\\sin x}{x} = \\int_0^{\\pi\/2} d\\theta \\frac{\\sin \\theta}{\\sin \\theta} = \\pi$$<\/p>\n<p>Therefore, the value of $$\\int_0^\\infty {\\frac{1}{{1 + {{\\text{x}}^2}}}} {\\text{dx}} + \\int_0^\\infty {\\frac{{\\sin {\\text{x}}}}{{\\text{x}}}} {\\text{dx}}$$ is $\\pi$.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Subscribe on 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